jdk carlson: exercises to atiyah and macdonald’s introduction to commutative algebra...

jdk carlson: Exercises to Atiyah and Macdonald’s Introduction to Commutative Algebra, 2019 palingenesis

revision of February 5, 2019

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Motivation

This note is intended to contain full solutions to all exercises in this venerable text,1 as well as proofs of resultsomitted or left to the reader. It has none of the short text’s pith or elegance, tending rather to the other extreme, citingchapter and verse from the good book and spelling out, step-by-step, things perhaps better left unsaid. It attempts toleave no “i” undotted, no “t” uncrossed, no detail unexplained. We are rather methodical in citing results when weuse them, even if they’ve likely long been assimilated by the reader. Having found some solution sets online unclearat points (sometimes due to our own shortcomings, at other times due to theirs), we in this note strive to sufferfrom the opposite problem. Often we miss the forest for the trees. We prefer to see this not as “pedantic,” but as“thorough.” Sometimes we have included multiple proofs if we have found them, or failed attempts at proof if theirfailure seems instructive. The work is our own unless explicitly specified otherwise. It is hoped that the prolix andoftentimes plodding nature of these solutions will illuminate more than it conceals.

1 [A–M]

2

Notation

Problems copied from the book and propositions are in italics, definitions emphasized (italic when surroundingtext is Roman, and Roman when surrounding text is italic), and headings in bold or italics, following the book, andour solutions and occasional comments in Roman. Solutions that were later supplanted by better ones but still mightpotentially be worth seeing have usually been included, but as footnotes, decreasing both page count and legibility.Propositions, exercises, theorems, lemmas, and corollaries from the main body of the text will always be cited as“(n.m)” or (n.m.t)”, where n is the chapter number, m the section number, and “t” an optional Roman numeral.For example, Proposition 1.10, part ii) is cited as “(1.10.ii)”. Propositions or assertions proved in these solutionsbut not stated in the text are numbered with an asterisk, e.g. “Proposition 4.12*” and thereafter “(4.12*)”. Exercisesfrom the “EXERCISES” sections that follow each chapter, on the other hand, we cite with (square) brackets as e.g.“[3.1]” and “[1.2.i]”. Displays in this document are referred back to as e.g. “Eq. 1.1” or “Seq. 2.2”. Ends of proofs forexercises go unmarked, though we will sometimes mark proofs for discrete propositions we prove in the course ofdoing problems or expanding upon material.

Our mathematical notation follows that of the book, with a few exceptions as noted below. For strict set con-tainment, “⊂” is supplanted by “(”, which is preferred for its lack of ambiguity; it is not to be confused with “6⊆”,which means “does not contain.” “3” is a backward “∈”, and means “contains the element” (rather than “such that”).The ideal generated by a set of elements is noted by listing them between parentheses: e.g., (x, y) is generated byx, y and (xα)α∈A is generated by xα : α ∈A. Contrastingly (and disagreeing with the book), we notate sequencesand ordered lists with angle brackets: ⟨x, y⟩ is an ordered pair and ⟨xα⟩α∈A is a list of elements xα indexed by a set A;in particular ⟨xn⟩n∈N is a sequence. Popular algebraic objects like N, Z,Q, R, and C will be denoted in blackboardbold instead of bold, and 0 is included in N. Fq denotes a (“the”) finite field with q elements. “a Ã A” means thata is an ideal of the ring A. For set complement/exclusion, the symbol “\” replaces “−” on the off chance it mightotherwise be confused with subtraction in cases (like topological groups) where both operations are feasible. Theset of units of the ring A will be denoted by A× rather than A∗, which is assigned a different meaning on p. 107.Bourbaki’s word “quasi-compact” for the condition that every open cover has a finite subcover (not necessarily re-quiring the space be Hausdorff) we replace with “compact”; this usage seems to hold generally outside of algebraicgeometry and most of the topologies we encounter here are not Hausdorff anyway. “ker”, “im”, and “coker” willgo uncapitalized. N(A) and R(A) always denote, respectively, the nilradical and the Jacobson radical of the ring A;we just write N and R where no ambiguity is possible. The notation idM : M →M for the identity map replaces thebook’s “1” as slightly more unambiguous; 1 or 1A is instead the unity of the ring A. “Multiplicative submonoid” ispreferred to the book’s “multiplicatively closed subset” as indicating that a subset of a ring contains 1 and is closedunder the ring’s multiplication. “Zorn’s Lemma,” capitalized, is the proper name of a result discovered by KazimierzKuratowski some thirteen years earlier than by Max Zorn. For a map f : A→ B , we can replace the arrow with “”when f is surjective, “” when it is injective, “,→” when it is an inclusion, and “

∼−→” when it is an isomorphism.A∼= B means that there exists some isomorphism between A and B , and X ≈ Y that X and Y are homeomorphictopological spaces. [M ] is the isomorphism class of M . For a map f : A→ B , if U ⊆ A and V ⊆ B are such thatf (U )⊆V , then f |VU is the restricted and corestricted map U →V . Very occasionally, κ, λ, µ may be cardinals (orhomomorphisms), and indices α, β, γ may be ordinals. ℵ0 is the cardinality of N.

3

Chapter 1: Rings and Ideals

Theorem 1.3. Every ring A 6= 0 has at least one maximal ideal.

In order to apply Zorn’s Lemma, it is necessary to prove that if ⟨aα⟩α∈A is a chain of ideals (meaning, recall, thatfor all α, β ∈A we have aα ⊆ aβ or aβ ⊆ aα) then the union a=

⋃

α∈A aα is an ideal. Indeed, if a, b ∈ a, then thereare α, β ∈A such that and a ∈ aα and b ∈ aβ. Without loss of generality, suppose aα ⊆ aβ. Then a, b ∈ aβ, so sinceaβ is an ideal we have a − b ∈ aβ ⊆ a. If x ∈ A and a ∈ a, then there is α ∈ A such that a ∈ aα. As aα is an ideal,xa ∈ aα ⊆ a. Therefore a is an ideal.

Exercise 1.12.i) a⊆ (a : b).

For each a ∈ a we have ab⊆ ab⊆ a, so a ∈ (a : b).

ii) (a : b)b⊆ a.

By definition, for x ∈ (a : b) we have xb⊆ a.

iii)

(a : b) : c

= (a : bc) =

(a : c) : b

.

x ∈

(a : b) : c

⇐⇒ xc⊆ (a : b) ⇐⇒ xcb⊆ a ⇐⇒ x ∈ (a : bc);

x ∈

(a : c) : b

⇐⇒ xb⊆ (a : c) ⇐⇒ xbc⊆ a ⇐⇒ x ∈ (a : bc).

iv)⋂

i ai : b

=⋂

i (ai : b).

x ∈

⋂

i

ai : b

⇐⇒ xb⊆⋂

i

ai ⇐⇒ ∀i (xb⊆ ai ) ⇐⇒ x ∈⋂

i

(ai : b).

v)

a :∑

i bi

=⋂

i (a : bi ).

x ∈

a :∑

i

bi

⇐⇒ a⊇ x

∑

i

bi

=∑

i

xbi ⇐⇒ ∀i (xbi ⊆ a) ⇐⇒ x ∈⋂

i

(a : bi ).

For an A-module M and subsets N ⊆ M and E ⊆A, define (N : E) := m ∈ M : Em ⊆N; for subsets N , P ⊆ Mand E ⊆A, define (N : P ) := a ∈A : aP ⊆N.

Note for future use that then ii) holds equally well for subsets a, b ⊆ M , or b ⊆ A and a ⊆ M ; iii) holds fora, b⊆M and c⊆A; and iv) and v) hold for modules a, ai and modules or ideals b, bi .

Exercise 1.13.−i) a⊆ b =⇒ r (a)⊆ r (b).

If x ∈ r (a), for some n > 0 we have xn ∈ a⊆ b, so x ∈ r (b).

0) r (an) = r (a) for all n > 0.

an ⊆ a, so by part −i) we have r (an) ⊆ r (a). If x ∈ r (a), then for some m > 0, x m ∈ a. But then x mn ∈ an andx ∈ r (an).

i) r (a)⊇ a.

4

Chapter 1: Rings and Ideals

For each a ∈ a we have a1 ∈ a, so a ∈ r (a).

ii) r

r (a)

= r (a).

x ∈ r

r (a)

⇐⇒ ∃n > 0

xn ∈ r (a)

⇐⇒ ∃n, m > 0

(xn)m = x mn ∈ a

⇐⇒ x ∈ r (a).

iii) r (ab) = r (a∩ b) = r (a)∩ r (b).

For the first equality, note (a∩b)2 ⊆ ab⊆ a∩b, so by parts 0) and−i), r (a∩b) = r

(a∩b)2

⊆ r (ab)⊆ r (a∩b).For the second, note that if m, n > 0 are such that x m ∈ a and xn ∈ b, then xmaxm, n ∈ a∩ b, and conversely.

iv) r (a) = (1) ⇐⇒ a= (1).

If r (a) = (1), then 1 ∈ r (a), so for some n we have 1= 1n ∈ a, and then a= (1).

v) r (a+ b) = r

r (a)+ r (b)

.

Since a, b ⊆ a+ b, by part −i) we have r (a), r (b) ⊆ r (a+ b), so r (a) + r (b) ⊆ r (a+ b). By parts −i), and ii),we see r

r (a) + r (b)

⊆ r

r (a+ b)

= r (a+ b). Conversely, by part i), we have a ⊆ r (a) and b ⊆ r (b), so adding,a+ b⊆ r (a)+ r (b). By part −i), r (a+ b)⊆ r

r (a)+ r (b)

.

vi) If p is prime, r (pn) = p for all n > 0.

By part 0), r (pn) = r (p). By (1.14), r (p) = p.

Proposition 1.17. iv*) Both extension and contraction are order-preserving with respect to containment; i.e. for idealsa1 ⊆ a2 of A we have ae

1 ⊆ ae2 and for ideals b1 ⊆ b2 of B we have bc

1 ⊆ bc2.1

If a1 ⊆ a2, then f (a1)⊆ f (a2), so ae1 = Bf (a1)⊆ Bf (a2) = ae

2. If b1 ⊆ b2, then bc1 = f −1(b1)⊆ f −1(b2) = bc

2.

Exercise 1.18. Let f : A→ B be a ring homomorphism, and let a, a j be ideals of A and b, b j ideals of B.∑

a j

e =∑

aej .

Because the homomorphism f distributes over finite sums and multiplication of ideals distributes over addition,

∑

a j

e= Bf

∑

a j

= B ·∑

f (a j ) =∑

Bf (a j ) =∑

aej .

(a1a2)e = ae

1ae2.

Because 1 ∈ B , we have B = BB ; as f is a homomorphism and ideal multiplication commutes,

(a1a2)e = Bf (a1a2) = BBf (a1) f (a2) = Bf (a1)Bf (a2) = ae

1ae2.

∑

b j

c ⊇∑

bcj .

Given any finitely many nonzero a j ∈ f −1(b j ), we have f∑

j a j

=∑

j f (a j ) ∈∑

j b j .

(b1b2)c ⊇ bc

1bc2.

If f (a j ) ∈ b j for j = 1, 2, then f (a1a2) = f (a1) f (a2) ∈ b1b2.

⋂

j a j

e ⊆⋂

j aej .

⋂

a j

e= Bf

⋂

a j

⊆ B ·⋂

f (a j ) =⋂

Bf (a j ) =⋂

aej .

⋂

b j

c =⋂

bcj .

⋂

b j

c= f −1

⋂

b j

=⋂

f −1(b j ) =⋂

bcj .

(a1 : a2)e ⊆ (ae

1 : ae2).

1 This is trivial, but the book never seems to explicitly state that it is the case, so here is a place to cite when we use it.

5

Ex. 1.1 Chapter 1: Rings and Ideals

By the result on multiplying extended ideals, and since (a1 : a2)a2 ⊆ a1 by (1.13.ii), we have

(a1 : a2)eae

2 =

(a1 : a2)a2

e (1.17.iv*)⊆ ae

1.

(b1 : b2)c ⊆ (bc

1 : bc2).

By the result on multiplying contracted ideals, and since (b1 : b2)b2 ⊆ b1 by (1.13.ii),

(b1 : b2)cbc

2 ⊆

(b1 : b2)b2

c (1.17.iv*)⊆ bc

1.

r (a)e ⊆ r (ae ).

Let b =∑

j b j f (x j ) for b j ∈ B and xn j

j ∈ a. Some f (x j )n j divides each term of b

∑

j (n j−1)+1, so b ∈ r (ae ).

r (b)c = r (bc ).

a ∈ f −1

r (b)

⇐⇒ ∃n > 0

f (an) = f (a)n ∈ b

⇐⇒ ∃n > 0

an ∈ f −1(b)

⇐⇒ a ∈ r

f −1(b)

.

The set of extended ideals is closed under sum and product, and the set of contracted ideals is closed under intersection,quotient, and radical.

It now suffices to show (bc1 : bc

2) = (bc e1 : bc e

2 )c . Indeed, using the preceding facts and (1.17.ii),

a ∈ (bc1 : bc

2) ⇐⇒ abc2 ⊆ bc

1 =⇒ f (a) f (bc2)⊆ f (bc

1) ⇐⇒ f (a)bc e2 ⊆ bc e

1 ⇐⇒ a ∈ (bc e1 : bc e

2 )c ;

a ∈ (bc e1 : bc e

2 )c ⇐⇒ f (a)bc e

2 ⊆ bc e1 =⇒ abc

2 ⊆ f −1 f (a)

bc2 = f −1 f (a)

bc ec2 ⊆ bc ec

1 = bc1 =⇒ a ∈ (bc

1 : bc2).

EXERCISES1. Let x be a nilpotent element of a ring A. Show that 1+ x is a unit of A. Deduce that the sum of a nilpotent element and a

unit is a unit.

The nilradical is a subset of the Jacobson radical, and by (1.9), for any x ∈R we have 1+ x = 1− (−x) a unit.2

Now if x is nilpotent and u a unit, then u−1x is nilpotent as well and u + x = u(1+ u−1x) is invertible.

2. Let A be a ring and let A[x] be the ring of polynomials in an indeterminate x, with coefficients in A. Let f = a0+ a1x+· · ·+ an xn ∈A[x]. Prove thati) f is a unit in A[x] ⇐⇒ a0 is a unit in A and a1, . . . , an are nilpotent.

⇐=: If a0 is a unit and the a j are nilpotent for j ≥ 1, then since N is an ideal by (1.7), the a j x j are nilpotent forj ≥ 1 and y =

∑nj=1 a j x j is nilpotent, and by [1.1], f = a0+ y is invertible.

=⇒: We induct on deg f . The degree zero case is trivial. Suppose we have proved the result for degrees < n, andlet deg f = n. Suppose that f is a unit, with inverse g =

∑mi=0 bi x i ; assume for uniformity of notation that a j = bi = 0

for integers i , j outside the ranges indicated. Write ck =∑k

j=0 a j bk− j for 0≤ k ≤ m+ n, so that 1= f g =∑

k ck xk .We have 1= c0 = a0b0, so a0 and b0 are units, and the other ck are all 0. Note in particular 0= cm+n = an bm : a powerof an annihilates bm . This is the r = 1 case of a general claim that a r

n bm+1−r = 0 for 1≤ r ≤ m+ 1. Indeed, fix suchan r and inductively assume a s

n bm+1−s = 0 for 1≤ s ≤ r . We have

0= cm+n−r = bm−r an + bm−r+1an−1+ · · ·+ bman−r .

Multiplying by a rn , we get

0= a r+1n bm−r + a r

n bm−r+1︸︷︷︸

0

(an−1an)+ · · ·+ an bm︸︷︷︸

0

(an−r a r−1n ),

2 Alternatively, let n > 0 be minimal such that xn = 0, and let y =∑n−1

j=0 (−x) j ; then

(1+ x)y =n−1∑

j=0

(−x) j −n∑

j=1

(−x) j = 1± xn = 1.

6

Chapter 1: Rings and Ideals Ex. 1.3

so 0 = a r+1n bm−r , completing the induction. When we get to r = m we see b0am+1

n = 0, so since b0 is a unit, an isnilpotent. Hence an xn is nilpotent, and by [1.1], f −an xn is a unit. This has degree< n, so by induction, a1, . . . , an−1are nilpotent.

ii) f is nilpotent ⇐⇒ a0, a1, . . . , an are nilpotent.

⇐=: Since N

A[x]

is an ideal by (1.7), if all a j ∈N, then all a j x j ∈N, so f =∑

a j x j ∈N.=⇒: On the other hand, for each prime pÃA, we have p[x]ÃA[x] prime since it is the kernel of the surjection

A[x] (A/p)[x], whose image is an integral domain by [1.2.iii]: if a j x j ∈ A[x] is a zero-divisor, there exists anonzero c ∈A with c ·

∑

a j x j = 0, so each c · a j = 0, and hence a j = 0 as A/p is an integral domain. Thus

N

A[x]

(1.8)⊆⋂

p[x]

=

⋂

p

[x] =N(A)[x].3

iii) f is a zero-divisor ⇐⇒ there exists a 6= 0 in A such that af = 0.

The “if” direction is trivial; the “only if” we prove by induction. We prove something slightly more specific: if anonzero g = b0+ b1x + · · ·+ bm x m ∈ (0 : f ) is of least possible degree m, then bm f = 0.

For the base case, if f = a0 has degree zero, then of course bma0 = bm f = 0. Suppose inductively that for allzero-divisors f ′ of degree n − 1 we know there is some b ∈ A such that b f ′ = 0. Let deg f = n and g and m be asbefore. Since f g = 0, also an bm = 0, so deg(an g )< m. As an g f = 0 as well, by minimality of m, we know an g = 0.Now 0= f g − an xn g = ( f − an xn)g . Since f ′ = f − an xn is of degree < n, by the inductive assumption bm f ′ = 0,so bm f = bman xn + bm f ′ = 0, completing the induction.

iv) f is said to be primitive if (a0, a1, . . . , an) = (1). Prove that if f , g ∈ A[x], then fg is primitive ⇐⇒ f and g areprimitive.

We can actually let A[x] := A[x1, . . . , xr ] be a polynomial ring in several indeterminates, writing x for the se-quence x1, . . . , xr , in the proof below.

Note that a polynomial is primitive just if no maximal ideal contains all its coefficients. Let mÃ A be maximal.Since A/m is a field, A[x]/m[x]∼= (A/m)[x] is an integral domain. Thus

f, g /∈m[x] ⇐⇒ f, g 6= 0 in (A/m)[x] ⇐⇒ f g 6= 0 in (A/m)[x] ⇐⇒ f g /∈m[x].

Therefore no maximal ideal contains all the coefficients of f g just if the same holds for f and g .4

This result is called Gauß’s Lemma and was originally proven in his Disquisitiones Arithmeticae for A= Z. Cf.also [9.2].

3. Generalize the results of Exercise 2 to a polynomial ring A[x1, . . . , xr ] in several indeterminates.

We start with an assumption about the ring B =A[x1, . . . , xr ] and prove the corresponding statement about thering B[y] =A[x1, . . . , xr , y] in one more indeterminate. For a multi-index α= ⟨ j1, . . . , jr ⟩we write aα, k := a j1, ..., jr , k

and xα := x j11 · · · x

jrr . If f ∈ B[y], we can write it as f =

∑

α, k aα, k xαyk =∑

k hk yk , where hk =∑

α aα, k xα ∈ B .

i) f is a unit in B[y] ⇐⇒ a0,0 is a unit in A and all other aα, k are nilpotent.

f ∈ B[y]×[1.2.i]⇐⇒B[y]/B

h0 ∈ B× and other hk ∈N(B)[1.3.i], [1.3.ii]⇐⇒B/A

a0,0 ∈A× and other aα, k ∈N(A).

3 Alternate proof: suppose f m = 0 and j is minimal with a j 6= 0. Then the lowest term amj x j m of f is 0, so a j is nilpotent and f − a j x j is

nilpotent. Repeatedly lopping off lowest terms, we see each a j ∈N(A).4 We also have the following generalization of the classical proof. Suppose f is not primitive, so that for some maximal m Ã A we have

f ∈m[x]. Then f g ∈m[x], so f g is not primitive. The same holds if g is not primitive.Now suppose f g is not primitive; we show one of f and g is also not. There is a maximal idealm containing all of the coefficients cr =

∑

j a j br− jof f g ; we suppose that neither all of the a j nor all of the bk lie inm and obtain a contradiction. There are a least J and a least K such that aJ , bK /∈m.Now m contains the coefficient cJ+K =

∑

j<J a j bJ+K− j + aJ bK +∑

k<K aJ+K−k bk , and each of the sums is in m by assumption, so aJ bK ∈m aswell. Since m is prime, we have aJ or bK in m, a contradiction.

7


Use of [1.3.ii] is permissible because it is independent, but we could also perform the induction in both exercises atthe same time.

ii) f is nilpotent ⇐⇒ all aα, k are nilpotent.

f ∈N

B[y] [1.2.ii]⇐⇒B[y]/B

all hk ∈N(B)[1.3.ii]⇐⇒B/A

all aα, k ∈N(A).

iii) f is a zero-divisor ⇐⇒ there exists a 6= 0 in A such that af = 0.

The inductive assumption will be that if g ∈ B is a zero-divisor, and b ∈ B is of minimal multidegree α (in thereverse lexicographic order) such that b g = 0, then if a is the coefficient of the leading term of b , we have ag = 0.

The “if” is again trivial. For the “only if,” suppose f is a zero-divisor in B[y]. By [1.2], there exists a nonzerob ∈ B such that b f = 0. Thus bhk = 0 for each k. By the inductive assumption the highest coefficient a ∈ A of b issuch that ahk = 0 for each k. Then af = 0.

iv) Prove for f , g ∈A[x1, . . . , xr ] that f g is primitive over A ⇐⇒ f and g are primitive over A.

The proof in [1.2.iv] goes through equally well in this case.

4. In the ring A[x], the Jacobson radical is equal to the nilradical.

We know N⊆R by (1.8), since maximal ideals are prime, so it remains to show all elements of R are nilpotent.Let f =

∑

a j x j ∈R, where a j ∈A. By (1.9), 1− x f is a unit. By [1.2.i], then, all a j ∈N, so by [1.2.ii], f ∈N.

5. Let A be a ring and let A[[x]] be the ring of formal power series f =∑∞

n=0 an xn with coefficients in A. Show thati) f is a unit in A[[x]] ⇐⇒ a0 is a unit in A.

⇐=: Supposing a0 is a unit, we construct an inverse g =∑

m bm x m to f . Let b0 = a−10 . We want f g =

∑

j c j x j = 1,

so for j ≥ 1 we want c j =∑ j

n=0 an b j−n = 0. Now suppose we have found satisfactory coefficients b j for j ≤ k. Weneed ck+1 = a0bk+1+

∑k+1n=1 an bk+1−n = 0; but we can solve this to find the solution bk+1 =−a−1

0

∑k+1

n=1 an bk+1−n

.Since we can do this for all k, we have constructed an inverse to f .=⇒ : If g =

∑

m bm x m is an inverse of f , then f g = 0 implies a0b0 = 1 so that a0 is a unit.

ii) If f is nilpotent, then an is nilpotent for all n ≥ 0. Is the converse true?

The two proofs of “=⇒” in [1.2.ii] both hold, mutatis mutandis, here.The converse is untrue. Let B = C [y1, y2, . . .] be a polynomial ring in countably many indeterminates over an

integral domain C , and let b= (y1, y22 , y3

3 , . . .) be the smallest ideal containing each ynn for n ≥ 1. Then writing zn for

the image of yn in A= B/b, we have znn = 0 and zn−1

n 6= 0. Thus an element of A is equal to zero just if, written as apolynomial in the zn over C , each term is divisible by some zn

n . Now let f =∑∞

n=0 zn xn ∈A[[x]]. By construction,each coefficient is nilpotent. However, for each n, one term of the coefficient in A of xn(n+1) in f n is zn

n+1, which isnonzero and cannot be cancelled, so f n 6= 0.

iii) f belongs to the Jacobson radical of A[[x]] ⇐⇒ a0 belongs to the Jacobson radical of A.

If the constant coefficient of g ∈A[[x]] is b0 ∈A, then the constant coefficient of 1− f g is 1− a0b0. Now

f ∈R

A[[x]] (1.9)⇐⇒ ∀g ∈A[[x]]

1− f g ∈A[[x]]× [1.5.i]⇐⇒ ∀b0 ∈A

1− a0b0 ∈A× (1.9)⇐⇒ a0 ∈R(A).

iv) The contraction of a maximal ideal m of A[[x]] is a maximal ideal of A, and m is generated by mc and x.

Since x ∈ A[[x]] has constant term 0 ∈R(A), by iii) above, x ∈R

A[[x]]

, and hence (x)⊆m. As m\(x) =mc ,we get m=mc +(x). Now A/mc ∼=A[[x]]/m is a field, so mc ÃA is maximal.5

5 I was tempted to use here that ae = aA[[x]] = a[[x]] for any aÃA, but it turns out this is wrong in general. It does hold for finitely generateda (see [Eisenbud, Ex. 7.13]), and it is true ([4.7.i]) that aA[x] = a[x] in A[x], but there are counterexamples if a is not finitely generated. Forexample, as in part ii) let C be a ring and b = (y1, y2, y3, . . .) in B = C [y1, y2, y3, . . .]. Then

∑

yn xn is in b[[x]] but not in bB[[x]]. To seethis, suppose not; then there is a finite collection of b j ∈ b and there are b ′j, n ∈ B such that

∑

yn xn =∑

j

b j∑

n b ′j, n xn, so that for all n wehave

∑

j b j b ′j, n = yn in B . But then b would be finitely generated by these finitely many b j , which is impossible because each b j is in a subring ofC generated over B by finitely many yn .

8


v) Every prime ideal of A is the contraction of a prime ideal of A[[x]].

Let pÃA be prime, and let q= pA[[x]]+ (x) be the ideal of A[[x]] generated by p and x. Evidently qc = p, andA[[x]]/q∼=A/p is an integral domain, so q is prime.

6. A ring A is such that every ideal not contained in the nilradical contains a nonzero idempotent (that is, an element e suchthat e2 = e 6= 0). Prove that the nilradical and Jacobson radical of A are equal.

N ⊆ R in any ring. Now suppose a /∈ N. Then (a) 6⊆ N, so there is a nonzero idempotent e = ax ∈ (a). Since(1− e)e = 0, we see 1− e is a zero-divisor, and hence not a unit; by (1.9), ax = e /∈R, so a /∈R. Thus R⊆N.

7. Let A be a ring in which every element satisfies xn = x for some n > 1 (depending on x). Show that every prime ideal inA is maximal.

Let p Ã A be prime and x ∈ A\p, with xn = x for some n ≥ 2. In the domain A/p we can cancel x 6= 0, and soxxn−2 = xn−1 = 1, showing an inverse x−1 = xn−2 exists. Thus A/p is a field and p was maximal.6

8. Let A be a ring 6= 0. Show that the set of prime ideals of A has minimal elements with respect to inclusion.

Partially order the set Spec(A) of prime ideals of A by p ≤ q :⇐⇒ p ⊇ q. We find minimal elements by Zorn’sLemma. Let ⟨pα⟩α∈A be a totally ordered chain in Spec(A), and let p =

⋂

α∈A pα be the intersection. It is an ideal.Now suppose xy ∈ p and x /∈ p. Then xy ∈ pα for each a; as pα is prime, this means each pα contains either x ory. Since x /∈ p, there is some β such that x /∈ pβ; since α ≥ β ⇐⇒ pα ⊆ pβ, we have for all α ≥ β that x /∈ pα, soy ∈ pα. As for γ ≤ β we have y ∈ pβ ⊆ pγ , we know y ∈ pα for all α ∈ A. Thus y ∈ p. Therefore p ∈ Spec(A), andSpec(A) has minimal elements.

9. Let a be an ideal 6= (1) in a ring A. Show that a= r (a) ⇐⇒ a is an intersection of prime ideals.

By (1.14), r (a) =⋂

p ∈ Spec(A) : a ⊆ p. If a is an intersection of prime ideals, it is the intersection r (a) of allprimes that contain it, and if not, it then cannot be r (a), so it must be a proper subset.

10. Let A be a ring, N its nilradical. Show the following are equivalent:i) A has exactly one prime ideal;ii) every element of A is either a unit or nilpotent;iii) A/N is a field.

i) =⇒ ii): If p is the only prime, N(1.8)= p and p is also the only maximal ideal. Then x /∈A×

(1.5)⇐⇒ x ∈ p=N.

ii) =⇒ iii): If x ∈A/N is nonzero, any lift x /∈N, and so has an inverse y. Then x−1 = y in A/N.

iii) =⇒ i): If A/N is a field, then N is a maximal ideal. Since every prime p⊇N, we have p=N the only prime.

11. A ring A is Boolean if x2 = x for all x ∈A. In a Boolean ring A, show thati) 2x = 0 for all x ∈A;

x + 1= (x + 1)2 = x2+ 12+ 2x = (x + 1)+ 2x, so subtracting x + 1 from both sides, 0= 2x.

ii) every prime ideal p is maximal, and A/p is a field with two elements;

By [1.7], every prime ideal is maximal. x2− x = x(x−1) = 0 holds for each x ∈A, so x(x−1) = 0 holds for eachx ∈A/p; as A/p is an integral domain, this means each element is either 0 or 1, so A/p∼= F2.

iii) every finitely generated ideal in A is principal.

We induct on the number of generators. The one-generator case is trivial. Suppose every ideal generated byn elements is principal, and a = (x1, . . . , xn , y). Let x generate (x1, . . . , xn), and let z = x + y − xy. Then xz =x2+ xy − x2y = x and yz = y, so a= (x, y) = (z).

6 Here is a more baroque proof. Since xn = x, if for any m > 0 we have x m = 0, then taking p such that n p > m, we see 0 = x m xn p−m =xn p= x, so N = (0). If 0 and 1 (possibly equal) are the only elements of A, we are done. If not, let x /∈ 0, 1. We have x(1− xn−1) = 0, and by

assumption x and xn−1 are nonzero, so 1− xn−1 is a zero-divisor, hence not a unit, and so xn−1 is not in the Jacobson radical by (1.9), meaningx is not in the Jacobson radical either. Thus R= (0).

9


12. A local ring contains no idempotent 6= 0, 1.

Let A be a ring. For any idempotent unit e , we have e = e−1e2 = e−1e = 1. Suppose e2 = e 6= 0, 1 in A. Then eis not a unit, and by (1.5) is contained in some maximal ideal m. Similarly (1− e)2 = 1− 2e + e2 = 1− e is anotheridempotent 6= 0, 1, hence not a unit. But were A local, e would be in m=R, so 1− e would be a unit by (1.9).7

13. Let K be a field and let Σ be the set of all irreducible monic polynomials f in one indeterminate with coefficients in K.Let A be the polynomial ring over K generated by indeterminates xf , one for each f ∈Σ. Let a be the ideal of A generatedby the polynomials f (xf ) for all f ∈Σ. Show that a 6= (1).

If a = (1), there exist finitely many af ∈A such that 1=∑

af f (xf ). The set I of xg occurring in this expression(not only those in the f (xf ), but also those occurring in the af ) is finite. We may enumerate I as x1, . . . , xi , . . . , xn ,corresponding to irreducible polynomials fi , and suppose n is minimal such that such an equation holds. WriteB =K[x1, . . . , xn−1], C = B[xn], and b :=

f1(x1), . . . , fn−1(xn−1)

Ã B . By minimality of n, the ideal b is proper, sobe = b[xn]ÃC is as well, while by the equation above, be +

fn(xn)

=C . Since b 6= B , we know B/b 6= 0. Let g bethe image of fn(xn) in (B/b)[xn]. Since fn is irreducible in K[xn], its degree degxn

fn ≥ 1 and also degxng ≥ 1. Then

0=C

be +

fn(xn)

∼=C/

b[xn]

(g )=

B[xn]/

b[xn]

(g )∼=(B/b)[xn](g )

6= 0,

which is a contradiction.8

Let m be a maximal ideal of A containing a and let K1 = A/m. Then K1 is an extension field of K in which each f ∈ Σhas a root. Repeat the construction with K1 in place of K, obtaining a field K2, and so on. Let L=

⋃∞n=1 Kn . Then L is a

field in which each f ∈Σ splits completely into linear factors. Let K be the set of all elements of L which are algebraic overK. Then K is an algebraic closure of K.

We should probably show that K is closed under subtraction and multiplication. Let a, b ∈ K have conjugatesai , b j over K . Then

∏

i , j (x − ai + b j ) is symmetric in the ai and the b j , and so has coefficients in K , so a− b ∈ K .Similarly

∏

i , j (x − ai b j ) is symmetric, so ab ∈K .

14. In a ring A, let Σ be the set of all ideals in which every element is a zero-divisor. Show that the setΣ has maximal elementsand that every maximal element of Σ is a prime ideal. Hence the set of zero-divisors in A is a union of prime ideals.

OrderΣ by inclusion; to show it has maximal elements, it suffices by Zorn’s Lemma to show every chain ⟨aα⟩α∈Ahas an upper bound inΣ. Let a=

⋃

α aα. It contains only zero-divisors, since if x ∈ a, then there is α such that x ∈ aα,and then by definition x is a zero-divisor.

Let p be a maximal element ofΣ; we must show it to be prime. Suppose x, y /∈ p. Then there are non-zero-divisorsin (x)+ p and (y)+ p, and their product is an element of (xy)+ p that is again a non-zero-divisor. Thus xy /∈ p, lestthere be something in p other than a zero-divisor. This shows p is prime.

ThusΣ has maximal elements and every element ofΣ is contained in one; considering principal ideals, this showsevery zero-divisor is in a maximal element of Σ. The last statement follows.

The prime spectrum of a ring15. Let A be a ring and let X be the set of all prime ideals of A. For each subset E of A, let V (E) denote the set of all prime

ideals of A which contain E. Prove thati) if a is the ideal generated by E, then V (E) =V (a) =V (r (a)).

Let p ∈ X. We have E ⊆ a, so if a ⊆ p, then E ⊆ p. On the other hand, if E ⊆ p, then a = AE ⊆ Ap = p. ThusV (E) =V (a). By (1.14), r (a) =

⋂

V (a), so p⊇ r (a) ⇐⇒ p⊇ a and V (a) =V

r (a)

.

ii) V (0) =X, V (1) =∅.For every prime ideal p we have 0 ∈ p and 1 /∈ p.

7 Cf. the implication iii) =⇒ ii) in [1.22] for another proof: by (1.4), each of (e) and (1− e) is contained in a maximal ideal, and we can showthe two are coprime, so no maximal ideal can contain both. Actually, there is even an isomorphism A

∼−→ A/(e)×A/(1− e), so A obviously hasmore than one maximal ideal.

8 This proof is taken from [Morandi].

10


iii) if ⟨Ei ⟩i∈I is any family of subsets of A, then V

⋃

i∈I

Ei

=⋂

i∈I

V (Ei ).

p ∈V

⋃

i

Ei

⇐⇒⋃

i

Ei ⊆ p ⇐⇒ ∀i ∈ I (Ei ⊆ p) ⇐⇒ ∀i ∈ I (p ∈V (Ei )) ⇐⇒ p ∈⋂

i

V (Ei ).

Note also for future use that⋃

Ei ⊆ p ⇐⇒⋃

AEi ⊆Ap= p ⇐⇒∑

AEi ⊆ p, so in particular for ideals ai wehave V

⋃

ai

=V∑

ai

iv) V (a∩ b) =V (ab) =V (a)∪V (b) for any ideals a, b of A.

Suppose ab⊆ p and b 6⊆ p. Then there is b ∈ b\p, and ab ∈ p for all a ∈ a, so the primality of p gives a ∈ p andthus a⊆ p. Thus if p ∈V (ab), we have shown a⊆ p or b⊆ p, so p ∈V (a)∪V (b). On the other hand, if p containsa or contains b, then it must contain the subset ab. Thus V (ab) =V (a)∪V (b).

Now ab⊆ a∩ b, so if a∩ b⊆ p, then ab⊆ p. On the other hand, if ab⊆ p, then we have shown either a⊆ p orb⊆ p, so since a∩ b is a subset of both of these we have a∩ b⊆ p. Thus V (a∩ b) =V (ab).

These results show that the sets V (E) satisfy the axioms for closed sets in a topological space. The resulting topology is calledthe Zariski topology. The topological space X is called the prime spectrum of A, and is written Spec(A).

16. Draw pictures of Spec(Z), Spec(R), Spec(C[x]), Spec(R[x]), Spec(Z[x]).

There is only one point, (0), in Spec(R).In Spec(Z), the elements are (0) and (p) for each positive prime p ∈N, and the closed sets are X , ∅, and all finite

sets containing (0).In C[x], all polynomials split into linear factors, so the irreducible polynomials are x −α for α ∈ C. Since C is

a field, this means the only primes are (0) and (x − α) for α ∈ C. The closed sets are again X , ∅, and all finite setscontaining (0). As a point set, it makes sense to think of X as the complex plane plus one additional dense point.

In R[x], all polynomials split into linear factors and polynomials of the form (x − α)(x + α) for α ∈ C withIm (α)> 0. Thus the primes of R[x] correspond to points of R, points of the upper half plane, and the dense point(0) again.

In Z[x], there are irreducible polynomials f of all degrees ≥ 1, giving rise to prime ideals ( f ), there are ideals(p) for all positive primes p ∈N, and there are ideals (p, f ) = (p) + ( f ), which are maximal. There is also (0). Theclosed sets are X , ∅, and all finite sets C containing (0) and such that if (p, f ) ∈C then (p), ( f ) ∈C .9

17. For each f ∈ A, let Xf denote the complement of V ( f ) in X = Spec(A). The sets Xf are open. These are called the basicopen sets of Spec(A). Show that they form a basis of open sets for the Zariski topology.

To see the collection Xf is a basis for the topology of X we can show it i) contains, for each p ∈Xf ∩Xg , an Xhwith p ∈Xh ⊆Xf ∩Xg , ii) includes ∅, and iii) covers X. These follow, respectively, from i), ii), and iii) below.

i) Xf ∩Xg =Xf g ;

Taking complements, this is the same as saying V ( f )∪V (g ) =V ( f g ), or that a prime contains f g if and only ifit contains one of f and g . But this is in the definition of a prime ideal.

ii) Xf =∅ ⇐⇒ f is nilpotent;

Xf =∅ ⇐⇒ V ( f ) =X ⇐⇒ ∀p ∈X ( f ∈ p)(1.8)⇐⇒ f ∈N.

iii) Xf =X ⇐⇒ f is a unit;

Xf =X ⇐⇒ V ( f ) =∅ ⇐⇒ ∀p ∈X ( f /∈ p)(1.5)⇐⇒ f ∈A×.

iv) Xf =Xg ⇐⇒ r

( f )

= r

(g )

;

9 Mumford’s famous picture of this space can be seen at [LeBruyn].

11


We can prove something slightly better: Xf ⊆Xg ⇐⇒ V (g )⊆V ( f ) ⇐⇒ r

( f )

⊆ r

(g )

. The first step is ob-vious because complementation is containment-reversing. Recall from [1.15.i] that V ( f ) =V

( f )

. For the secondstep, we generalize again, from ( f ), (g )ÃA to arbitrary ideals a, b.10 Note the antitone Galois correspondence11

a⊆⋂

Y ⇐⇒ ∀p ∈ Y (a⊆ p) ⇐⇒ Y ⊆V (a)

between ideals aÃA and subsets Y ⊆ Spec(A). Applying it to Y =V (b) yields

V (b)⊆V (a) ⇐⇒ a⊆⋂

V (b)(1.14)= r (b)

(1.13)⇐⇒ r (a)⊆ r (b).

v) X is compact (that is, every open covering of X has a finite sub-covering).

This follows from the more general vi), taking f = 1.

vi) More generally, each Xf is compact.

Recall that the closed sets of X are all of the form V (a) for aÃ A. Let a collection

X \V (aα)

αof open sets be

given. This collection covers Xf if and only if the following equivalent conditions hold:⋂

α

V (aα)⊆V ( f )[1.15]⇐⇒ V

∑

α

aα

⊆V

( f ) proof⇐⇒of iv)

r

( f )

⊆ r

∑

α

aα

(1.13)⇐⇒ ∃m ≥ 1

h

f m ∈∑

α

aα

i

;

and then our task is to find a finite set of aα for which the last still holds. But each element of∑

aα is a finite sum ofelements from the individual aα, so f m ∈

∑

aα just if for some finite subset a j nj=1 we have f m∈∑n

j=1 a j .

vii) An open subset of X is compact if and only if it is a finite union of sets Xf .

Each Xf is compact, and a union of a finite collection of compact sets is compact12, so a finite union of basic opensets Xf is compact.

On the other hand, suppose a set is open and compact. Since it is open, we can write it as a union of some basicopen sets X fα

; since it is compact, we can take a finite subcover, showing it is a union of finitely many basic opensets.

18. For psychological reasons it is sometimes convenient to denote a prime ideal of A by a letter such as x or y when thinkingof it as a point of X = Spec(A). When thinking of x as a prime ideal of A, we denote it by px (logically, of course, it is thesame thing). Show thati) The set x is closed (we say that x is a “closed point”) in Spec(A) ⇐⇒ px is maximal;

Let Y ⊆X, and let V (a)⊆X be a closed set. Recall our Galois correspondence: Y ⊆V (a) ⇐⇒ a⊆⋂

y∈Y py .

Y =⋂

n

V (a) : Y ⊆V (a)o

=⋂

n

V (a) : a⊆⋂

y∈Y

py

o

[1.15]= V

∑

n

a : a⊆⋂

y∈Y

py

o

=V

⋂

y∈Y

py

, (1.1)

so x is closed just if x=V (px ), or in other words iff no other prime contains px .

ii) x=V (px );

xEq.=1.1

V

⋂

px

=V (px ).

10 The proof branches here; if you like, you can apply what follows in this footnote instead of the final two sentences in the body text. Recallalso from (1.14) that for all ideals aÃA we have r (a) =

⋂

V (a), from [1.15.i] that V (a) =V

r (a)

, and from [1.15.iii] that V (−) is containment-reversing. Finally note taking intersections of collections is a containment-reversing operation. Then

r (a)⊆ r (b) =⇒ V (b) =V

r (b)

⊆V

r (a)

=V (a);

V (b)⊆V (a) =⇒ r (a) =⋂

V (a)⊆⋂

V (b) = r (b).

11 [WPGalois]12 To see this, let a cover V of the finite union K =

⋃nj=1 of compact sets K j be given. For each K j take a finite subcollectionU j ⊆V covering

K j ; then⋃

j U j ⊆V is a finite collection covering K .

12


iii) y ∈ x ⇐⇒ px ⊆ py ;

y ∈ x ii)=V (px ) ⇐⇒ px ⊆ py .

iv) X is a T0-space (this means that if x, y are distinct points of X, then either there is a neighborhood of x which does notcontain y, or else there is a neighborhood of y which does not contain x).

If every neighborhood of x contains y and vice versa, then y ∈ x and x ∈ y, so by part iii); px ⊆ py ⊆ px andx = y.

19. A topological space X is said to be irreducible if X 6= ∅ and if every pair of non-empty open sets in X intersect, orequivalently if every non-empty open set is dense in X. Show that Spec(A) is irreducible if and only if the nilradical of Ais a prime ideal.

∅ is not an intersection of two nonempty open sets just if X is not a union of two proper closed sets V (a), V (b).By the proof of [1.17.iv],

X =V (0) 6=V (a) ⇐⇒ r (a) 6⊆ r (0)(1.8)= N

(1.13)⇐⇒ a 6⊆N,

and by [1.15.iv], V (a)∪V (b) = V (ab), so X is irreducible just if for all ideals a, b 6⊆ N we also have ab 6⊆ N; bycontraposition, this is that ab⊆N =⇒ a⊆N or b⊆N.

But this condition is just a rephrasing of primality: if N is prime, then as in [1.15.iv], ab ⊆ N =⇒ a ⊆ N orb⊆N; conversely, if the condition holds, then ab ∈N =⇒ (a)(b )⊆N =⇒ (a) or (b )⊆N =⇒ a or b ∈N, so Nis prime.

20. Let X be a topological space.i) If Y is an irreducible (Exercise 19) subspace of X, then the closure Y of Y in X is irreducible.

Let open U , V ⊆ Y be non-empty. Then U ∩Y and V ∩Y are non-empty by the definition of closure. Since Yis irreducible, U ∩V ∩Y 6=∅, and a fortiori U ∩V 6=∅.

ii) Every irreducible subspace of X is contained in a maximal irreducible subspace.

We apply Zorn’s Lemma. Order the irreducible subspaces Σ of X by inclusion, and let ⟨Yα⟩ be a linearly orderedchain. Set Z =

⋃

αYα; we will be done if we can show Z ∈Σ. Let U , V ⊆ Z be open and non-empty. By definition,U meets some Yα and V meets some Yβ. Without loss of generality, suppose α≤β. Then as Yα ⊆ Yβ, we have bothU ∩Yβ and V ∩Yβ non-empty and open in Yβ, so by irreducibility of Yβ we have U ∩V ∩Yβ 6=∅ and U ∩V 6=∅,showing Z is irreducible.

iii) The maximal irreducible subspaces of X are closed and cover X. They are called the irreducible components of X.What are the irreducible components of a Hausdorff space?

To see that a maximal irreducible subspace is closed, note that its closure is irreducible by part i) and containsit, and so equals it by maximality. To see the maximal irreducible subspaces cover X, note that each singleton isirreducible and contained in some maximal irreducible subspace.

Every subspace of a Hausdorff space is Hausdorff. If a Hausdorff space has two distinct points, they have twodisjoint neighborhoods by definition, so the space is not irreducible. Thus the irreducible components of a Hausdorffspace are the singletons.

iv) If A is a ring and X = Spec(A), then the irreducible components of X are the closed sets V (p), where p is a minimalprime ideal of A (Exercise 8).

Any closed subset of X is of the form V (a) for some ideal aÃA, and is homeomorphic to Spec(A/a) by [1.21.iv].By [1.19], V (a) = V

r (a)

is irreducible if and only if N(A/a) is prime, or equivalently if r (a) is prime in A; sothe irreducible closed subspaces of X are V (p) for p ∈ X . Such a V (p) is maximal just if there is no q ∈ X withV (p)(V (q), or equivalently, by the proof of [1.17.iv], there is no prime q( p.

13


21. Let φ : A→ B be a ring homomorphism. Let X = Spec(A) and Y = Spec(B). If q ∈ Y , then φ−1(q) is a prime ideal ofA, i.e., a point of X. Hence φ induces a mapping φ∗ : Y →X. Show thati) If f ∈A then φ∗−1(Xf ) = Yφ( f ), and hence that φ∗ is continuous.

q ∈ Yφ( f ) ⇐⇒ φ( f ) /∈ q ⇐⇒ f /∈φ−1(q) ⇐⇒ φ∗(q) =φ−1(q) ∈Xf ⇐⇒ q ∈ (φ∗)−1(Xf ).

ii) If a is an ideal of A, then φ∗−1

V (a)

=V (ae ).

First, note that (1.17.i) implies extension and contraction of ideals form an isotone Galois correspondence:13

a⊆ bc ⇐⇒ ae ⊆ b.

Indeed, if a⊆ bc , then extending, ae ⊆ bc e ⊆ b, and if ae ⊆ b, then contracting, a⊆ aec ⊆ bc . Now for q ∈ Spec(B),

q ∈ (φ∗)−1V (a)

⇐⇒ φ∗(q) ∈V (a) ⇐⇒ a⊆ qc ⇐⇒ ae ⊆ q ⇐⇒ q ∈V (ae ).

iii) If b is an ideal of B, then φ∗(V (b)) =V (bc ).

By Eq. 1.1 from [1.18.i], φ∗(V (b)) is the set of prime ideals containing⋂

φ∗

V (b)

, which ideal equals

⋂

qc : b⊆ q ∈ Y (1.18)=

⋂

b⊆q∈Y

qc=(1.14)

r (b)c(1.18)= r (bc ).

But V

r (bc )

=V (bc ).

iv) If φ is surjective, then φ∗ is a homeomorphism of Y onto the closed subset V

ker(φ)

of X. (In particular, Spec(A)and Spec(A/N) (where N is the nilradical of A) are naturally homeomorphic.)

If φ : A→ A/a is surjective, (1.1) gives an containment-preserving and -reflecting bijection between the set ofideals b Ã A containing a and the ideals b/a Ã A/a. Since this relation preserves and reflects primes (p. 9), for anyideal b/a of A/a,

φ∗(V

b/a)

=

p ∈ Spec(A) : b/a⊆ p/a ∈ Spec(A/a)

=V (b),

so φ∗|V (a) is a bijection taking closed sets to closed sets, continuous by part i), and hence a homeomorphism. Forthe parenthetical remark, note that Spec(A) =V

N(A)

by (1.8).

v) If φ is injective, then φ∗(Y ) is dense in X. More precisely, φ∗(Y ) is dense in X ⇐⇒ ker(φ)⊆N.

The second statement does imply the first, because if φ is injective, then indeed ker(φ) = 0⊆N. Now φ∗(Y ) isdense just if

X =φ∗(Y ) =φ∗

V (0) [1.12.iii]= V (0c ) =V

ker(φ)

.

But ker(φ) is contained in every prime of A if and only if it is contained in their intersection, which by (1.8) is N(A).

vi) Let ψ : B→C be another ring homomorphism. Then (ψ φ)∗ =φ∗ ψ∗.

By definition, a ∈ (ψ φ)∗(px ) ⇐⇒ ψ

φ(a)

∈ px ⇐⇒ φ(a) ∈ψ∗(px ) ⇐⇒ a ∈φ∗

ψ∗(px )

.

vii) Let A be an integral domain with just one nonzero prime ideal p, and let K be the field of fractions of A. Let B =(A/p)×K. Define φ : A→ B by φ(x) = (x, x), where x is the image of x in A/p. Show that φ∗ is bijective but not ahomeomorphism.

Ahas two prime ideals, p and (0), and B , a product of two fields, has prime ideals q1 = 0×K and q2 = (A/p)×0;the zero ideal of B is not prime. Now φ∗(q1) = x ∈ A : x = 0 = p, and φ∗(q2) = x ∈ A : x = 0 = (0), so φ∗ isbijective. However, by [1.18.iii] p ∈ (0) in Spec(A), while in Spec(B) we have q2 /∈ q1= q1 (both primes beingmaximal), so φ∗ cannot be a homeomorphism.

13 [WPGalois]

14


22. Let A=∏n

i=1 Ai be a direct product of rings Ai . Show that Spec(A) is the disjoint union of open (and closed) subspacesXi , where Xi is canonically homeomorphic with Spec(Ai ).

Let pri : A→Ai be the canonical projection, and bi =∏

j 6=i Aj its kernel; then by [1.21.iv], pr∗i is a homeomor-phism Spec(Ai )

≈−→V (bi ). Since⋂n

i=1 bi = 0, it follows by [1.15.iv] that⋃

V (bi ) =V⋂

bi

=V (0) = Spec(A), sothe V (bi ) cover Spec(A).14 Since bi + b j = A for i 6= j and by [1.15], V (bi )∩V (b j ) = V (bi + b j ) = V (1) = ∅, itfollows the V (b j ) are disjoint. Since the complement

⋃

j 6=i V (b j ) of each V (bi ) is a finite union of closed sets, theV (bi ) are also open.

Conversely, let A be any ring. Show that the following statements are equivalent:i) X = Spec(A) is disconnected.ii) A∼=A1×A2 where neither of the rings A1, A2 is the zero ring.iii) A contains an idempotent 6= 0, 1In particular, the spectrum of a local ring is always connected (Exercise 12).

We showed ii) =⇒ i) above; in this case X =X1qX2 is a disjoint union two non-empty open sets.

i) =⇒ iii): Suppose X = Spec(A) is disconnected; then by definition it is a disjoint union of two non-emptyclosed sets V (a), V (b). By [1.15.iii] we have ∅=V (a)∩V (b) =V (a+b), so no prime contains a+b, which mustthen equal (1). Let a ∈ a and b ∈ b be such that a + b = 1. By [1.15.iv], X = V (a) ∪V (b) = V (ab), so by (1.8),ab ⊆N and there is some n ≥ 1 with (ab )n = 0. We have (1) = (an) + (b n) by (1.16), so we can find e ∈ (an) suchthat 1− e ∈ (b n). We then have e− e2 = e(1− e) ∈ (ab )n = 0, so e = e2. If e = 1 we would have 1 ∈ a and if e = 0 wewould have 1 ∈ b, contrary to assumption, so e is a nontrivial idempotent.

iii) =⇒ ii): Suppose e 6= 0, 1 is an idempotent. Then as in the proof of [1.12], 1− e is also an idempotent 6= 0, 1,and neither is a unit. This means (e) and (1−e) are proper, nonzero ideals, and they are coprime since e+[1−e] = 1.Since (e)(1− e) = (e − e2) = (0), then, (1.10.i) shows (e)∩ (1− e) = (0). Let φ : A→A/(e)×A/(1− e) be the naturalhomomorphism. (1.10.ii,iii) show φ is an isomorphism.15

23. Let A be a Boolean ring (Exercise 11), and let X = Spec(A).i) For each f ∈A, the set Xf (Exercise 17) is both open and closed in X.

Xf is open because it is the complement of V ( f ). It is open because V ( f ) =X1− f . Indeed, X =Xf ∪X1− f , sinceany ideal containing both f and 1− f contains 1; and Xf ∩X1− f =Xf (1− f ) =X0 =∅ by [1.17.i] and [1.17.ii].

ii) Let f1, . . . , fn ∈A. Show that X f1∪ · · · ∪X fn

=Xf for some f ∈A.⋃

Xfi=⋃

X \V ( fi )

= X \⋂

V ( fi ) = X \V∑

( fi )

by [1.15]. By [1.11.iii],∑

( fi ) = ( f ) for some f ∈ A, so⋃

Xfi=X \V ( f ) =X f .

iii) The sets Xf are the only open subsets of X which are both open and closed.

Let U be both open and closed. Since it is closed and X is compact, U is compact. By [1.17.vii], U is a union offinitely many X f j

. By part ii), it is an Xf for some f ∈A.

iv) X is a compact Hausdorff space.

The compactness of X is [1.17.v]. To show X is Hausdorff, let x, y ∈ X ; we will show that if they do not havedisjoint neighborhoods Xf and X1− f , then x = y. Now Xf 3 x and X1− f 3 y just if f /∈ px and 1− f /∈ py . By part i),this is the same as saying f /∈ px and f ∈ py . If no such f exists, we have py ⊆ px , and since [1.11.ii] showed py ismaximal, px = py .

14 In unnecessary detail, for each j there is an element e j ∈A with a 1 in the j coordinate and 0 at each other coordinate. Let aÃA be an ideal,and a = ⟨a1, . . . , an⟩ ∈ a. Then ae j = a j e j ∈ a, and a =

∑

a j e j =∑

ae j , so a=∑

ae j . We have b j =∑

i 6= j (ei ). To see Spec(A) =⋃

X j , note thatif a Ã A contains neither ei nor e j for some i 6= j , then since ei e j = 0 ∈ a, we know a is not prime. Therefore all the prime ideals of A containsome b j , and thus are of the form p= pr∗j (p j ) = p j e j + b j for some p j ∈ Spec(Aj ), and hence are in some X j .

15 We can also show ii) =⇒ iii): ⟨1, 0⟩ ∈A1×A2 is an idempotent 6= 0, 1.

15


24. Let L be a lattice, in which the sup and inf of two elements a, b are denoted by a∨b and a∧b respectively. L is a Booleanlattice (or Boolean algebra) ifi) L has a least element and a greatest element (denoted by 0, 1 respectively);ii) each of ∨, ∧ is distributive over the other;iii) Each a ∈ L has a unique “complement” a′ ∈ L such that a ∨ a′ = 1 and a ∧ a′ = 0.For example, the set of all subsets of a set, ordered by inclusion, is a Boolean lattice.

Let L be a Boolean lattice. Define addition and multiplication in L by the rules

a+ b = (a ∧ b ′)∨ (a′ ∧ b ), ab = a ∧ b .

Verify that in this way L becomes a Boolean ring, say A(L).

To verify the ring axioms, we first require some lemmas about Boolean algebra.

• Commutativity: The supremum x ∨ y, by definition, is the unique z ≥ x, y such that for all other w ≥ x, ywe have w ≥ z, and this definition is symmetric in x and y; thus x ∨ y = y ∨ x. Dually, x ∧ y = y ∧ x is theunique z ≤ x, y such that for all other w ≤ x, y we have w ≤ z, and this definition is symmetric in x and y.

• Associativity: (x∨y)∨ z is the unique least element t ≥ x∨y, z. Then we have t ≥ x, y, z, so (x∨y)∨ z ≥ x∨y∨z, the joint supremum of x, y, z. On the other hand, x∨y∨z ≥ x, y, z as well, so x∨y∨z ≥ x∨y, z, and bydefinition x∨y∨z ≥ (x∨y)∨z. Since we have both inequalities and ⟨L, ≤⟩ is a partial order, (x∨y)∨z = x∨y∨z.Symmetrically, x ∨ y ∨ z = x ∨ (y ∨ z). The proof (x ∧ y)∧ z = x ∧ y ∧ z = x ∧ (y ∧ z) is dual, exchanging ∧ for∨ and ≥ for ≤.

• Idempotence: x ∨ x = x = x ∧ x, for x is the least element greater than both x and x, and also the greatestelement less than both of them.

• Absorption: x ∨ (x ∧ y) = x = x ∧ (x ∨ y), because x is the least element≥ x, x ∧ y and the greatest≤ x, x ∨ y.

• Identity: for all x ∈ L we have 0≤ x ≤ 1, so 0∧ x = 0 and 0∨ x = x = x ∧ 1 and x ∨ 1= 1.

• De Morgan’s laws: (x ∨ y)′ = x ′ ∧ y ′ and (x ∧ y)′ = x ′ ∨ y ′. For the first, note that

(x ′ ∧ y ′)∧ (x ∨ y) = (x ′ ∧ y ′ ∧ x)∨ (x ′ ∧ y ′ ∧ y) = (0∧ y ′)∨ (x ′ ∧ 0) = 0∨ 0= 0,

(x ′ ∧ y ′)∨ (x ∨ y) = (x ′ ∨ x ∨ y)∧ (y ′ ∨ x ∨ y) = (1∨ y)∧ (x ∨ 1) = 1∧ 1= 1;

since (x ∨ y)′ is postulated to be unique with these properties, we have (x ∨ y)′ = x ′ ∧ y ′. The other law(x ∧ y)′ = x ′ ∨ y ′ is dual; the proof is the same, exchanging ∧ for ∨ and vice versa everywhere.

From now on write · for ∧. We prove a few more miscellaneous facts.

• a+ b = (ab ′)∨ (a′b ) = (a ∨ a′)(a ∨ b )(b ′ ∨ a′)(b ′ ∨ b ) = 1(a ∨ b )(a′ ∨ b ′)1= (a ∨ b )(a′ ∨ b ′).

• (a+ b )′ =

(a ∨ b )(a′ ∨ b ′)′ = (a ∨ b )′ ∨ (a′ ∨ b ′)′ = a′b ′ ∨ ab .

• 1′ = 0: for 1∧ 0= 1 and 1∧ 0= 0.

• 1+ a = (a ∨ 1)(a′ ∨ 0) = 1a′ = 1∧ a′ = a′.

Now we can prove the ring axioms for A(L).

• Commutativity of +: a+ b = (a ∨ b )(a′ ∨ b ′) = (b ∨ a)(b ′ ∨ a′) = b + a.

• Associativity of ·: · is ∧.

• Commutativity of ·: · is ∧.

16


• Associativity of +: (a+ b )+ c is

[a+ b ]∨ c

[a+ b ]′ ∨ c ′

= (ab ′ ∨ a′b ∨ c)(a′b ′ ∨ ab ∨ c ′)

= ab ′a′b ′ ∨ ab ′ab ∨ ab ′c ′∨a′ba′b ′ ∨ a′bab ∨ a′b c ′∨ca′b ′ ∨ cab ∨ cc ′ (1.2)

= 0∨ 0∨ ab ′c ′ ∨ 0∨ 0∨ a′b c ′ ∨ ca′b ′ ∨ cab ∨ 0.

= ab ′c ′ ∨ a′b c ′ ∨ a′b ′c ∨ ab c .

Write x+ = x and x− = x ′. Then (a+ b )+ c is the supremum of the four possible terms a±b±c± with an oddnumber of+’s. This is invariant under permutations of a, b , c , so by commutativity, a+(b+c) = (b+c)+a =(a+ b )+ c .

• 1a = a: for 1a = 1∧ a = a.

• a+ 0= a: for a+ 0= (a ∨ 0)(a′ ∨ 1) = a1= a.

• a =−a: for a+ a = (a ∨ a)(a′ ∨ a′) = aa′ = 0.

• Distributivity of · over +: a(b + c) = a

[bc ′]∨ [b ′c]

= abc ′ ∨ ab ′c , while

ab + ac = (ab ∨ ac)

[ab ]′ ∨ [ac]′

= ab [ab ]′ ∨ ac[ab ]′ ∨ ab [ac]′ ∨ ac[ac]′

= 0∨ ac[a′ ∨ b ′]∨ ab [a′ ∨ c ′]∨ 0

= aca′ ∨ acb ′ ∨ aba′ ∨ abc ′

= 0∨ abc ′ ∨ 0∨ ab ′c = abc ′ ∨ ab ′c .

• Boolean ring: a2 = a ∧ a = a by idempotence of ∧.

Conversely, starting from a Boolean ring A, define an ordering on A as follows: a ≤ b means that a = ab . Show that, withrespect to this ordering, A is a Boolean lattice. In this way we obtain a one-to-one-correspondence between (isomorphismclasses of) Boolean rings and (isomorphism classes of) Boolean lattices.

Write L = L(A) for ordered set. We verify the partial order axioms, the lattice axioms, and the Boolean algebraaxioms.

• ≤ is reflexive: a = aa, so a ≤ a.

• ≤ is antisymmetric: Suppose a ≤ b ≤ a. Then a = ab and b = ab , so a = b .

• ≤ is transitive: Let a ≤ b ≤ c . Then a = ab and b = bc , so a = ab = a(b c) = (ab )c = ac and a ≤ c .

• Binary suprema exist in ⟨L, ≤⟩: Let a∨ b = a+ b +ab . We have a(a∨ b ) = a(a+ b +ab ) = a2+ab +a2b = aby [1.11.i], so a ≤ a ∨ b , and symmetrically b ≤ a ∨ b . Now suppose a, b ≤ c . Then a = ac and b = bc , so(a ∨ b )c = (a+ b + ab )c = ac + bc + a(b c) = a+ b + ab = a ∨ b , and a ∨ b ≤ c . This shows a ∨ b is the leastupper bound of a, b in the partial order ⟨L, ≤⟩.

• Binary infima exist in ⟨L, ≤⟩: Let a∧b = ab . Then (a∧b )a = aba = ab = a∧b , so a∧b ≤ a, and symmetricallya ∧ b ≤ b . Now suppose c ≤ a, b . Then c = ca = cb , so c(a ∧ b ) = c(ab ) = (ca)b = cb = c , and so c ≤ a ∧ b .This shows a ∧ b is the greatest lower bound of a, b in the partial order ⟨L, ≤⟩.

• A least element exists in ⟨L, ≤⟩: For any a ∈A we have 0= 0a, so 0≤ a.

• A greatest element exists in ⟨L, ≤⟩: For any a ∈A we have a = a1, so a ≤ 1.

• ∧ distributes over∨: (a∨b )∧c = (a+b+ab )c = ac+bc+abc = ac+bc+(ac)(b c) = ac∨bc = (a∧c)∨(b∧c).

17


• ∨ distributes over ∧: (a ∧ b )∨ c = ab ∨ c = ab + c + abc , while also

(a ∨ c)∧ (b ∨ c) = (a+ c + ac)(b + c + bc)= ab + ac + abc + cb + cc + cbc + acb + acc + acbc= ab + 2ac + 3abc + 2bc + c= ab + c + abc .

• Each a ∈ A has a unique complement: Suppose a′ is such that a ∨ a′ = 1 and a ∧ a′ = 0. Then aa′ = 0, while1= a+ a′+ aa′ = a+ a′. Then a′ = 1− a = 1+ a. Thus the complement, if it exists, is unique. And a′ = 1+ ais indeed a complement: a ∧ a′ = a(1+ a) = a+ a2 = 0, while a ∨ a′ = a+ a′+ aa′ = a+(1+ a)+ 0= 1.

We finally should verify that these correspondences are inverse. Let A be a Boolean ring, and A′ =A

L(A)

. Then·A′ = ∧L(A) = ·A, and addition in A′ is

a+A′ b = ab ′ ∨ a′b = ab ′+A a′b +A ab ′a′b = ab ′+A a′b = a(1+A b )+A (1+A a)b = a+A ab +A b +A ab = a+A b ,

so A

L(A)

=A. On the other hand let L be a Boolean algebra and L′ = L

A(L)

. Then ∧L′ = ·A(L) = ∧L. Finally thejoin in L′ is given by

x ∨L′ y = x + y + xyEq.=1.2

xyxy ∨L x ′y ′xy ∨L x ′y(xy)′ ∨L xy ′(xy)′

=xy ∨L 0∨L x ′y(x ′ ∨L y ′)∨L xy ′(x ′ ∨L y ′)

=xy ∨L x ′yx ′ ∨L x ′yy ′ ∨L xy ′x ′ ∨L xy ′y ′

=xy ∨L x ′y ∨L xy ′

=xy ∨L xy ′ ∨L yx ∨L yx ′

=x(y ∨L y ′)∨L y(x ∨L x ′)=x ∨L y

25. From the last two exercises deduce Stone’s theorem, that every Boolean lattice is isomorphic to the lattice of open-and-closedsubsets of some compact Hausdorff topological space.

Let a Boolean algebra L be given, and let A be the associated Boolean ring. By [1.23.iv], X = Spec(A) is a com-pact Hausdorff space. Let B be the algebra of simultaneously open and closed sets in X. By the definition of openand closed, it is closed under binary union and intersection, so it is a sublattice of the power set ⟨P (X ), ⊆⟩ underinclusion. By the definition of a topology, ∅, X ∈ B . By set algebra, ∪ and ∩ each distribute over the other. Thecomplement V of an open and closed set U is open and closed, and is the unique V ⊆P (X ) with U ∪V = X andU ∩V =∅. Thus B is a Boolean algebra.

By [1.23.i] and [1.23.iii], B is the set of Xf for f ∈A, so the correspondenceφ : L→ B taking f 7→Xf is surjective.Since no nonzero element of A is nilpotent, by [1.17.ii], φ is also injective.

To showφ is an order isomorphism (hence a Boolean algebra isomorphism), it remains to show that f ≤ g ⇐⇒Xf ⊆ Xg . Now f ≤ g ⇐⇒ f = f g =⇒ f ∈ (g ), and conversely if f = ag ∈ (g ), then f g = (ag )g = ag = f . Thusf ≤ g ⇐⇒ f ∈ (g ). Now

f ∈ (g ) ⇐⇒ ∃n ≥ 1

f = f n ∈ (g )

⇐⇒ f ∈ r

(g ) (1.13)⇐⇒ r

( f )

⊆ r

(g )

,

and from the proof of [1.17.iv], r

( f )

⊆ r

(g )

⇐⇒ Xf ⊆ Xg . Therefore f ≤ g ⇐⇒ Xf ⊆ Xg , so φ is an orderisomorphism.

26. Let A be a ring. The subspace of Spec(A) consisting of the maximal ideals of A, with the induced topology, is called themaximal spectrum of A and is denoted by Max(A). For arbitrary commutative rings it does not have the nice functorialproperties of Spec(A) (see Exercise 21), because the inverse image of a maximal ideal under a ring homomorphism neednot be maximal.

Let X be a compact Hausdorff space and let C (X ) denote the ring of all real-valued continuous functions on X (addand multiply functions by adding and multiplying their values). For each x ∈ X, let mx be the set of all f ∈ C (X ) suchthat f (x) = 0. The ideal mx is maximal, because it is the kernel of the (surjective) homomorphism C (X )→R which takesf to f (x). If eX denotes Max

C (X )

, we have therefore defined a mapping µ : X → eX, namely x 7→mx .We shall show that µ is a homeomorphism of X onto eX.

18


i) Let m be any maximal ideal of C (X ), and let V =V (m) be the set of common zeros of the functions in m: that is,

V = x ∈X : f (x) = 0 for all f ∈m.

Suppose that V is empty. Then for each x ∈X there exists fx ∈m such that fx (x) 6= 0. Since fx is continuous, there is anopen neighborhood Ux of x in X on which fx does not vanish. By compactness a finite number of the neighborhoods,say Ux1

, . . . , Uxn, cover X. Let

f = f 2x1+ · · ·+ f 2

xn.

Then f does not vanish at any point of X, hence is a unit in C (X ). But this contradicts f ∈m, hence V is not empty.Let x be a point of V . Then m⊆mx , hence m=mx because m is maximal. Hence µ is surjective.

ii) By Urysohn’s lemma (this is the only non-trivial fact required in the argument) the continuous functions separate thepoints of X. Hence x 6= y =⇒ mx 6=my , and therefore µ is injective.

iii) Let f ∈C (X ); letUf = x ∈X : f (x) 6= 0

and leteUf = m ∈ eX : f /∈m.

Show that µ(Uf ) = eUf . The open sets Uf (resp. eUf ) form a basis of the topology of X (resp. eX ) and therefore µ is ahomeomorphism.Thus X can be reconstructed from the ring of functions C (X ).

For each f ∈C (X ), we have

x ∈Uf ⇐⇒ f (x) 6= 0 ⇐⇒ f /∈mx ⇐⇒ mx ∈ eUf ,

so µ restricts to a bijection Uf ↔ eUf . It remains to show these sets form bases.The Uf will form a basis for X if whenever x ∈W ⊆X with W open, there is an f ∈C (X ) such that x ∈Uf ⊆W .

But as X is compact Hausdorff, it is normal, and so the Urysohn lemma applies to show closed sets can be separatedby continuous functions. Thus there is f ∈ C (X ) such that f (X \W ) = 0 and f (x) = 1, and then evidentlyx ∈Uf ⊆W .

Each eUf is open the subspace topology inherited from Spec

C (X )

, being the intersection of eX with the open setSpec

C (X )

f of [1.17]. As these sets form a basis for Spec

C (X )

(see [1.17]), the eUf form a basis for eX.

Affine algebraic varieties27. Let k be an algebraically closed field and let

fα(t1, . . . , tn) = 0

be a set of polynomial equations in n variables with coefficients in k. The set X of all points x = ⟨x1, . . . , xn⟩ ∈ kn whichsatisfy these equations is an affine algebraic variety.

Consider the set of all polynomials g ∈ k[t1, . . . , tn] with the property that g (x) = 0 for all x ∈X. This set is an idealI (X ) in the polynomial ring, and is called the ideal of the variety X . The quotient ring

P (X ) = k[t1, . . . , tn]/I (X )

is the ring of polynomial functions on X, because two polynomials g , h define the same polynomial function on X if andonly if g − h vanishes at every point of X, that is, if and only if g − h ∈ I (X ).

Let ξi be the image of ti in P (X ). The ξi (1≤ i ≤ n) are the coordinate functions on X: if x ∈X, then ξi (x) is the ithcoordinate of x. P (X ) is generated as a k-algebra by the coordinate functions, and is called the coordinate ring (or affinealgebra) of X.

As in Exercise 26, for each x ∈X let mx be the ideal of all f ∈ P (X ) such that f (x) = 0; it is a maximal ideal of P (X ).Hence, if eX =Max

P (X )

, we have defined a mapping µ : X → eX, namely x 7→mx . It is easy to show that µ is injective:if x 6= y, we must have xi 6= yi for some i (1≤ i ≤ n), and hence ξi − xi is in mx but not in my , so that mx 6=my . Whatis less obvious (but still true) is that µ is surjective. This is one form of Hilbert’s Nullstellensatz (see Chapter 7).

Abbreviate k[t ] := k[t1, . . . , tn]. As in [1.26], mx is maximal because it is the kernel of the surjective homomor-phism f 7→ f (x) : P (X )→ k. To be more explicit about what mx looks like, note that if x = ⟨x1, . . . , xn⟩, then the

19


polynomial function ξi−xi ∈ P (X ) vanishes at x, so that ξi−xi ∈mx . On the other hand, since (t1−x1, . . . , tn−xn)is the kernel of the surjective homomorphism k[t ] k taking 1 7→ 1 and ti 7→ xi , it is a maximal ideal of k[t ], soby the correspondence (1.1) we have (ξ1− x1, . . . , ξn − xn)⊆mx maximal, which shows the the containment mustin fact be an equality.

We then want to show the images of these mx are the only maximal ideals of P (X ). By (1.1), it will suffice to dothis for X = kn and P (X ) = k[t ] and then prove that x ∈ X ⇐⇒ I (X ) ⊆ mx , which will be item 8 in a list ofremarks that follows.

First we show all maximal ideals of P (X ) come from points. One way is to use an equivalent result traditionallycalled the weak Nullstellensatz; see [5.17] for a statement and proof. Another is to use Zariski’s Lemma ((5.24), [5.18],(7.9)) that any field L finitely generated as an algebra over a field K is a finite algebraic extension of K, which impliesboth. This is done in [5.19]. Here is a more elementary proof16 avoiding the technology of Chapter 5, but it tooruns through Zariski’s Lemma.

Lemma 1.27.1*. If an integral domain Ais algebraic over a field k, then Ais a field.

Proof. Let 0 6= a ∈ A. Since A is a domain and a is algebraic over k, the kernel of the projection k[x] k[a] is anonzero prime ideal. But k[x] is a PID, so this kernel is maximal, k[a] is a field, and a is a unit.

Proposition 1.27.2*. If k ⊆ L are fields in some integral domain B finitely generated as a k-algebra, L is algebraic overk.

Proof. Imagine there were a transcendental element a ∈ L, and include a in a finite set T of generators for B asa k-algebra. Select a maximal k-algebraically independent set S ⊆ T containing a, so that the field of fractions L′

of B is a finite extension of k(S).17 Picking an k(S)-basis of L′ gives us a representation φ of multiplication on L′

by square matrices over k(S). Writing the entries of the matrices φ(t ) for t ∈ T as fractions in k[S], let g be theproduct of all the denominators, so that φ(B) has entries in k[S, g−1]. If p ∈ k[a] is any irreducible element, thenp−1 ∈ L since L is a field, so φ(p−1) is a diagonal matrix with entries p−1. Then this entry lies in k[S, g−1], so thereis some positive power g m such that g m p−1 ∈ k[S], meaning p|g m . Since p is irreducible and k[a] and k[S], beingisomorphic to polynomial rings, are UFDs, it follows p is a scalar multiple of one of those finitely many irreduciblefactors qi ∈ k[S] of g which also lie in k[a]. By unique factorization, it follows each element of k[a] is divisible bya qi ; but this obviously doesn’t hold of 1+

∏

qi , so we have a contradiction.

Taking B = L in (1.27.2*) yields Zariski’s Lemma. One could at this point use the proof of [5.19], but we willfollow an alternate route, proving another lemma we will encounter again later.

Lemma 1.27.3*. If k is a field, B a finitely generated k-algebra, and A→ B a k-algebra homomorphism, then contractionsin Aof maximals ideal of B are maximal.

Proof. Since B is a finitely generated k-algebra, it is a fortiori finitely generated over A, so B/m is a field finitelygenerated over the integral domain A/mc . By (1.27.2*), B/m is algebraic over k. Since A/mc is contained in B/m, itis also algebraic over k too. But then A/mc is a field by (1.27.1*), so mc is maximal.

Proposition 1.27.4*. If k is an algebraically closed field, all maximal ideals of the polynomial ring k[t ] are of the formmx .

Proof. Let m Ã k[t ] be maximal. Then each contraction m ∩ k[ti ] is maximal in the PID k[ti ] by (1.27.3*), andhence generated by an irreducible polynomial.18 As k is algebraically closed, this polynomial is linear, so we mayrescale it to be ti − xi for some xi ∈ k. But then ti − xi ∈m, so mx ⊆m and m=mx .

We take this opportunity to collect some remarks, culminating in the desired item 8. If S ⊆ k[t ] is a set ofpolynomials, generating the ideal a= (S), we define the zero set Z(S) of S to be

Z(S) = Z(a) :=n

⟨a1, . . . , an⟩ ∈ kn : ∀f ∈ a

f (a1, . . . , an) = 0

o

;

16 [Kemper, Prop. 1.5]17 This proof can be modified to use concepts from later on: pick a finite basis for L′ over k(S); each basis element satisfies a monic polynomial

with coefficients in k(S). Let g ∈ k[S] be a common multiple of the denominators of these coefficients. Then L′ is integral over k[S]g by (5.3),so k[S]g is a field by (5.7) or [5.5.i]. But this is a contradiction, for the polynomial ring k[S] cannot be a Goldman domain by (5.18.1*).

18 Note that the only reason we need the lemmas is to show this polynomial is not zero.

20


with this definition,19 it is clear that an affine algebraic variety X is just a set Z(S) for some dimension n and someS ⊆ k[t ]; and the maximal ideals of P (kn) = k[t ] referred to above are mx = I

x

.We have the following inclusion relations20 involving Z and I , for x ∈ kn , X , X1, X2 ⊆ kn , and S, S1, S2 ⊆ k[t ].

0. X ⊆ Z(S) ⇐⇒ S ⊆ I (X ): for both mean f (x) = 0 for all f ∈ S and all x ∈X .

1. X1 ⊆X2 =⇒ I (X2)⊆ I (X1): for if f ∈ k[t ] vanishes on X2, it also does on the subset X1.

2. S1 ⊆ S2 =⇒ Z(S2)⊆ Z(S1): for if S2 annihilates x ∈ kn , so does the subset S1.

3. X ⊆ ZI (X ): “X is annihilated by everything annihilating X ”; apply item 0 to I (X )⊆ I (X ).

4. S ⊆ I Z(S): “S vanishes on everything S vanishes on”; apply item 0 to Z(S)⊆ Z(S).

5. I (X ) = I ZI (X ): for X3.⊆ ZI (X ), so I ZI (X )

1.⊆ I (X ); but I (X )

4.⊆ I ZI (X ).

6. Z(S) = ZI Z(S): for S4.⊆ I Z(S), so ZI Z(S)

2.⊆ Z(S); but Z(S)

3.⊆ ZI Z(S).

7. Z(mx ) = x: if I

x

=mx annihilates y ∈ kn , then ∀i (yi − xi = 0), so y = x;

and x3.⊆ Z(mx ).

8. X = Z(S) =⇒

I (X )⊆mx ⇐⇒ x ∈X

: if I (X )⊆mx , then x 7.= Z(mx )2.⊆ ZI (X ) = ZI Z(S) 6.= Z(S) =X ;

and if x ∈X, then I (X )1.⊆mx .

9. r

I (X )

= I (X ): Since N(k) = 0, if f (x)m = 0 for m ≥ 1, then f (x) = 0.

Note as a consequence of item 8 that a variety is non-empty just if its ideal is contained in some mx . In particular,if there were some maximal ideal m that were not one of the mx , we would have Z(m) =∅.

28. Let f1, . . . , fm be elements of k[t1, . . . , tn]. They determine a polynomial mapping φ : kn → k m : if x ∈ kn , the coordi-nates of φ(x) are f1(x), . . . , fm(x).

Let X, Y be affine algebraic varieties in kn , k m respectively. A mapping φ : X → Y is said to be regular if φ is therestriction to X of a polynomial mapping from kn to k m .

If η is a polynomial function on Y , then η φ is a polynomial function on X. Hence φ induces a k-algebra homo-morphism P (Y )→ P (X ), namely η 7→ η φ. Show that in this way we obtain a one-to-one correspondence between theregular mappings X → Y and the k-algebra homomorphisms P (Y )→ P (X ).

The map φ# induced by φ is a ring homomorphism: for all x ∈X, η, ζ ∈ P (Y ), and ∗ ∈ +, −, ·, we have

(η ∗ ζ ) φ

(x) = (η ∗ ζ )

φ(x)) = η

φ(x)

∗ ζ

φ(x)

=

(η φ) ∗ (ζ φ)

(x).

Preservation of k is trivial: for a ∈ k we have a φ= a because a takes no arguments.To see that the correspondenceφ 7→φ# is injective, supposeφ# =ψ# forφ, ψ regular mappings X → Y induced,

respectively, by coordinates φ j , ψ j : kn → k for 1 ≤ j ≤ m. Letting υi (1 ≤ i ≤ m) be the coordinate functions onY , we then have

φi = υi φ=φ#(υi ) =ψ

#(υi ) = υi ψ=ψi

on X, so φi =ψi on X ; thus φi −ψi ∈ I (X ) for each i , and φ=ψ as regular maps from X.To see the correspondence is surjective, let λ be a k-algebra homomorphism P (Y )→ P (X ). Precomposing with

a quotient projection π : k[u] := k[u1, . . . , um] → P (Y ) from the polynomial ring, we have a homomorphismκ : k[u]→ P (X ). By the universal property of polynomial rings, this function is uniquely determined by its valueson indeterminates, say φi = κ(ui ) ∈ P (X ), which each define a regular function X → k. Let φ : X → k m be theregular function with coordinates φi . Then if η ∈ k[u], we have

κ(η) = η

κ(u1), . . . , κ(um)

= η(φ1, . . . , φm) = η φ.

19 This is closely related indeed to the sets V (E) of [1.15], to the extent that the same letter V is often used, and the topology on kn gottenby taking the Z(S) as closed sets is also called the Zariski topology. The correspondence is gotten by taking A= k[t ]; then under the bijectionkn ↔Max(A) (to be proved), we have Z(S)↔V (S)∩Max(A).

20 The first three show among other things show Z and I form an antitone Galois connection between the powersets P

k[t ]

and P (kn), aspartially ordered by inclusion—see e.g. [WPGalois]—and the next four are formal consequences of this Galois connection.

21


To show λ = φ# is as hoped, it remains to show that imφ ⊆ Y . This is the case just if for all η ∈ I (Y ) ⊆ k[u] andx ∈X we have η

φ(x)

= 0, or in other words if ηφ= 0; but this is true because ηφ= κ(η) = λ

π(η)

= λ(0) = 0.For later use, note that given X φ−→ Y ψ−→ Z we have (ψ φ)# = (φ# ψ#). Indeed, if ζ ∈ P (Z), then

(ψ φ)#(ζ ) = ζ ψ φ=φ#(ζ ψ) =φ#ψ#(ζ )

. (1.2)

This shows the coordinate ring is a contravariant functor from the category of affine algebraic varieties and regularmaps to the category of finitely generated k-algebras and k-algebra homomorphisms. This functor is actually anequivalence of categories.

Note in particular that in this framework, point inclusions 0 → x ,→ X correspond bijectively to k-algebrahomomorphisms P (X )→ k.

22

Chapter 2: Modules

Exercise 2.2. i) Ann(M +N ) =Ann(M )∩Ann(N ).

a ∈Ann(M +N ) ⇐⇒ 0= a(M +N ) = aM + aN ⇐⇒ aM = aN = 0 ⇐⇒ a ∈Ann(M )∩Ann(N ).

ii) (N : P ) =Ann

(N + P )/N

.

xP ⊆N ⇐⇒ x(N + P ) = xN + xP ⊆N ⇐⇒ x

(N + P )/N

= 0 in (N + P )/N .

Proposition 2.9. i) LetM ′ u−→M v−→M ′′→ 0 (2.1)

be a sequence of A-modules and homomorphisms. Then Seq. 2.1 is exact ⇐⇒ for all A-modules N , the sequence

0→Hom(M ′′, N ) v−→Hom(M , N ) u−→Hom(M ′, N ) (2.2)

is exact.

The book proves that if Seq. 2.2 is exact for all N , then Seq. 2.1 is exact. So suppose Seq. 2.1 is exact, let N beany A-module, and consider Seq. 2.2. Let φ ∈Hom(M ′′, N ). If 0 = φ v = v(φ), then φ = 0 since v is surjective;thus v is injective. We have u

v(φ)

= v(φ) u = φ v u = φ 0 = 0, so u v = 0. Now let ψ ∈ Hom(M , N )and suppose ψ ∈ ker u. Then ψ u = 0, so ker v = im u ⊆ kerψ, and ψ factors through the quotient moduleM ′′ = im v ∼=M/u(M ′) (see p. 19): there is ψ ∈Hom(M ′′, N ) with ψ= ψ v = v(ψ). Thus ψ ∈ im v.

ii) Let0→N ′ u−→N v−→N ′′ (2.3)

be a sequence of A-modules and homomorphisms. Then Seq. 2.3 is exact ⇐⇒ for all A-modules M , the sequence

0→Hom(M ,N ′) u−→Hom(M ,N ) v−→Hom(M ,N ′′) (2.4)

is exact.

Suppose Seq. 2.3 is exact, let M be an A-module, and consider Seq. 2.4. Let φ ∈ Hom(M , N ′). If φ ∈ ker u,then u φ = 0; since u is injective, φ = 0. Also v

u(φ)

= v u φ = 0 φ = 0, so v u = 0. Finally, letψ ∈ ker v ⊆Hom(M , N ). Then v ψ= 0, so imψ⊆ ker v = im u. Since u is injective, φ= u−1 φ is a well-definedmap such that u(φ) = u φ=φ.

For the other direction, suppose that for all A-modules M , Seq. 2.4 is exact, and consider Seq. 2.3. First, letM = Z. Suppose n′ ∈N ′ is such that u(n′) = 0. Let φ : Z→N ′ be given by φ(1) = n′. Then (u φ)(1) = u(n′) = 0,so φ ∈ ker u; since u is injective, φ= 0, so n′ = 0. This shows u is injective. Also, letting M = N ′, and consideringthe map idN ′ , we get 0 = v

u(idN ′)

= v u idN ′ = v u. Finally, let n ∈ ker v ⊆ N , and let ψ : Z→ N be givenby ψ(1) = n. Then im

v(ψ)

= im(v ψ) = v(nZ) = 0. By exactness of Seq. 2.4, there is ψ : Z → N ′ such thatu(ψ) = u ψ=ψ; in particular, n =ψ(1) = u

ψ(1)

, so n ∈ im u, and ker v ⊆ im u.

23

Chapter 2: Modules

Proposition 2.10. The Snake Lemma.

Starting with the two middle rows exact and maps α, β, γ making the two middle squares commute, we derivethe rest of the commutative diagram. For convenience, we have renamed objects and maps and will tend to omitparentheses. Except where otherwise noted, all deductions are by commutativity or exactness.

kerα _iα

κ′ // kerβ _iβ

λ′ // kerγ _iγ

δ

//

A

α

κ // B

β

λ // C

γ

// 0

0 // D

qα

φ // E

qβ

ψ // F

qγ

cokerαφ // cokerβ

ψ // cokerγ

• Note that if a ∈ kerα, then αa = 0, so βκa = φαa = 0, andκa ∈ kerβ; thus we can define a restriction κ′ : kerα→ kerβ.

• In the same way, we can define a restriction λ′ : kerβ→ kerγ .

• Since φ(imα) = im(βκ), φ induces a map φ : cokerα →cokerβ taking d = d + imα 7→φd + imβ=φd .

• ψ : cokerβ→ cokerγ is defined similarly.

The new squares all commute by definition.

• We check that the connecting map δ := qαφ−1βλ−1iγ is well defined. Let c ∈ kerγ and b ∈ λ−1c. Since

ψβb = γλb = γc = 0, there is a unique d ∈ D such that φd = βb , so δ can assign the value d = d + imαto c . We made a choice of b in this definition. Suppose b ′ ∈ λ−1c as well, and d ′ is the unique element ofφ−1βb ′. Since λ(b − b ′) = c − c = 0, there is a ∈A with κa = b − b ′. As

φd =βb =β(b ′+κa) =βb ′+φαa =φ(d ′+αa),

from the injectivity of φ we see d = d ′+αa, so that d = d ′ and δc is well defined.

Now we show exactness of kerα→ kerβ→ kerγ → cokerα→ cokerβ→ cokerγ . First come the easier parts:

• iγλ′κ′ = λκiα = 0iα = 0. As iγ is a monomorphism, λ′κ′ = 0.

• ψφqα = qγψφ= qγ0= 0. As qα is an epimorphism, ψφ= 0.

• δλ′ = qαφ−1β(λ−1iγλ

′) = qαφ−1(βiβ) = qαφ

−10= 0.

• φδ= (φqα)φ−1βλ−1iγ = (qβφ)φ

−1βλ−1iγ = (qββ)λ−1iγ = 0λ−1iγ = 0.

The other containments aren’t much worse:

• kerλ′ ⊆ imκ′: Suppose b ∈ kerβ∩kerλ. Then there is a ∈ A such that κa = b . Now φαa =βκa =βb = 0,and since φ is injective, a ∈ kerα; thus b ∈ imκ′.

• kerψ ⊆ imφ: Suppose ψe = 0; then there is c ∈ C such that ψe = γ c . As λ is surjective, there is b ∈ Bsuch that λb = c , and then ψβb = γλb = γc = ψe , so e −βb ∈ kerψ = imφ. Letting a ∈ A be such thatφa = e −βb , we see φa = e −βb = e , so e ∈ imφ.

• kerδ ⊆ imλ′: Suppose 0= δc and pick b ∈ λ−1c. Then d = 0 for the unique d ∈φ−1βb, so d ∈ imα. Ifαa = d , then βκa =φαa =φd =βb , so b −κa ∈ kerβ. We have λ(b −κa) = c − 0, so c ∈ imλ′.

• kerφ⊆ imδ: Suppose φ(d ) =φd = 0, so that there exists b ∈β−1φd. Setting λb = c , we see γc = γλb =ψβb = (ψφ)d = 0, so c ∈ kerγ . Now δc = qαφ

−1βλ−1c = qαφ−1βb = qαφ

−1φd = d , so d ∈ imδ.

Further, if the first sequence given is short exact (i.e., κ is injective), then κ′, as a restriction of κ, is injective; and ifthe second sequence given is short exact (i.e., ψ is surjective), then ψ is surjective, for if f ∈ cokerγ , there is somee ∈ E such that ψe = f , and then ψe = f .

24

Chapter 2: Modules

Proposition 2.12. Let M1, . . . , Mr be A-modules. Then there exists a pair ⟨T , g ⟩ consisting of an A-module T and anA-multilinear mapping g : M1× · · ·×Mr → T with the following property:

Given any A-module P and any A-multilinear mapping f : M1×· · ·×Mr → P, there exists a unique A-homomorphismf ′ : T → P such that f ′ g = f .

Moreover, if ⟨T , g ⟩ and ⟨T ′, g ′⟩ are two such pairs with this property, then there exists a unique isomorphism j : T →T ′ such that j g = g ′.

“The details may safely be left to the reader.” First we prove existence. Let C be the free A-module on M =∏r

j=1 M j , and let D be the submodule generated by the following elements, for all x j , x ′j ∈ M j ( j = 1, . . . , r ), anda ∈A:

⟨x1, . . . , x j + x ′j , . . . , xr ⟩− ⟨x1, . . . , x j , . . . , xr ⟩− ⟨x1, . . . , x ′j , . . . , xr ⟩,

⟨x1, . . . , ax j , . . . , xr ⟩− a⟨x1, . . . , x j , . . . , xr ⟩.

Let T = C/D , and write the image of (x1, . . . , xr ) ∈ C as x1 ⊗ · · · ⊗ xr ∈ T ; these elements evidently generateT , and we have, for all x j , x ′j ∈M j ( j = 1, . . . , r ), and a ∈A,

x1 ⊗ · · · ⊗ (x j + x ′j ) ⊗ · · · ⊗ xr = (x1 ⊗ · · · ⊗ x j ⊗ · · · ⊗ xr )+ (x1 ⊗ · · · ⊗ x ′j ⊗ · · · ⊗ xr , )

x1 ⊗ · · · ⊗ ax j ⊗ · · · ⊗ xr = a(x1 ⊗ · · · ⊗ x j ⊗ · · · ⊗ xr ),

so the map g : M → T given by ⟨x1, . . . , xr ⟩ 7→ x1 ⊗ · · · ⊗ xr is A-multilinear.Let P be an A-module; any map f : M → P extends uniquely to a linear map f : C → P since C is freely generated

over A by the elements of M . f is A-multilinear just if f vanishes on the elements we specified to generate D andthus induces on the quotient a unique A-linear map f ′ : T → P taking x1 ⊗ · · · ⊗ xr 7→ f (x1, . . . , xr ); in this case,f = f ′ g .

Now suppose ⟨T ′, g ′⟩ has the same universal property as ⟨T , g ⟩. Then the A-multilinear map g : M → T factorsthrough g ′ as g = j g ′ for a unique A-linear map j : T ′→ T . Similarly, g ′ : M → T ′ factors through g as g ′ = j ′ gfor a unique j ′ : T → T ′, and we have

g = j g ′ = j j ′ g .

Because g : M → T is itself multilinear, by the definition of a tensor product, there is a unique linear map i : T → Tsuch that g = i g ; but clearly the identity map idT has this property, so j j ′ = idT . Symmetrically,

g ′ = j ′ g = j ′ j g ′,

so j ′ j = idT ′ and j and j ′ = j−1 are isomorphisms.

Proposition 2.14. Let M , N , P be A-modules. Then there exist the following unique isomorphisms:i) M ⊗N →N ⊗M , x ⊗ y 7→ y ⊗ x;

The map ⟨x, y⟩ 7→ ⟨y, x⟩ 7→ y ⊗ x from M ×N → N ×M → N ⊗M is bilinear, so it corresponds to a uniquelinear map M ⊗N → N ⊗M taking x ⊗ y 7→ y ⊗ x. The same argument provides a map N ⊗M → M ⊗N takingy ⊗ x 7→ x ⊗ y. These maps are clearly inverse on the decomposable elements x ⊗ y ∈ M ⊗N and y ⊗ x ∈ N ⊗M ,and since these elements generate the modules, the maps are inverse.

ii) (M ⊗N )⊗ P →M ⊗ (N ⊗ P )→M ⊗N ⊗ P, (x ⊗ y)⊗ z 7→ x ⊗ (y ⊗ z) 7→ x ⊗ y ⊗ z;

The book demonstrates the isomorphism (M ⊗ N ) ⊗ P → M ⊗ N ⊗ P . A symmetric argument provides anisomorphism M ⊗ (N ⊗ P )→M ⊗N ⊗ P .

iii) (M ⊕N )⊗ P → (M ⊗ P )⊕ (N ⊗ P ), ⟨x, y⟩⊗ z 7→ ⟨x ⊗ z, y ⊗ z⟩;

Let f : (M ⊕N )×P → (M ⊗P )⊕ (N ⊗P ) be given by f

⟨x, y⟩, z

:= ⟨x⊗ z, y⊗ z⟩. We claim it is A-bilinear. Let

25

Chapter 2: Modules

x, x ′ ∈M, y, y ′ ∈N, z, z ′ ∈ P, and a ∈A: then indeed

f

a⟨x, y⟩, z

= f

⟨ax, ay⟩, z

= ⟨ax ⊗ z, ay ⊗ z⟩=

a(x ⊗ z), a(y ⊗ z)

= a⟨x ⊗ z, y ⊗ z⟩= af

⟨x, y⟩, z

,

f

⟨x, y⟩, az

= ⟨x ⊗ az, y ⊗ az⟩=

a(x ⊗ z), a(y ⊗ z)

= a⟨x ⊗ z, y ⊗ z⟩= af

⟨x, y⟩, z

,

f

⟨x, y⟩+ ⟨x ′, y ′⟩, z

= f

⟨x + x ′, y + y ′⟩, z

=

(x + x ′)⊗ z, (y + y ′)⊗ z

= ⟨x ⊗ z, y ⊗ z⟩+ ⟨x ′⊗ z, y ′⊗ z⟩= f

⟨x, y⟩, z

+ f

⟨x ′, y ′⟩, z

,

f

⟨x, y⟩, z + z ′

=

x ⊗ (z + z ′), y ⊗ (z + z ′)

= ⟨x ⊗ z + x ⊗ z ′, y ⊗ z + y ⊗ z ′⟩= ⟨x ⊗ z, y ⊗ z⟩+ ⟨x ⊗ z ′, y ⊗ z ′⟩= f

⟨x, y⟩, z

+ f

⟨x, y⟩, z ′

.

Thus f factors through the canonical map g : (M ⊕N )× P → (M ⊕N )⊗ P as f = f ′⊗ g , where

f ′: (M ⊕N )⊗ P → (M ⊗ P )⊕ (N ⊗ P ),⟨x, y⟩⊗ z 7→ ⟨x ⊗ z, y ⊗ z⟩

is A-linear.On the other hand, we also have bilinear maps

j1: M × P → (M ⊕N )⊗ P,⟨x, z⟩ 7→ ⟨x, 0⟩⊗ z

andj2 : N × P → (M ⊕N )⊗ P,⟨y, z⟩ 7→ ⟨0, y⟩⊗ z,

which give rise to linear maps

1: M ⊗ P → (M ⊕N )⊗ P,x ⊗ z 7→ ⟨x, 0⟩⊗ z

and2 : N ⊗ P → (M ⊕N )⊗ P,

y ⊗ z 7→ ⟨0, y⟩⊗ z.

By the universal property of the direct sum, we get a unique A-linear map

j : (M ⊗ P )⊕ (N ⊗ P )→ (M ⊕N )⊗ P,⟨x ⊗ z, 0⟩ 7→ ⟨x, 0⟩⊗ z,⟨0, y ⊗ z⟩ 7→ ⟨0, y⟩⊗ z.

Now the image of ⟨x, y⟩⊗ z ∈ (M ⊕N )⊗ P under j f is

j (x ⊗ z, y ⊗ z) = ⟨x, 0⟩⊗ z + ⟨0, y⟩⊗ z = ⟨x, y⟩⊗ z,

and since these elements generate (M ⊕N )⊗ P, we see j f is the identity. Similarly, the images of the elements⟨x ⊗ z, 0⟩ and ⟨0, y ⊗ z⟩ of (M ⊗ P )⊕ (N ⊗ P ) under f j are respectively

f

⟨x, 0⟩⊗ z

= ⟨x ⊗ z, 0⟩ and f

⟨0, y⟩⊗ z

= ⟨0, y ⊗ z⟩,

and these elements generate (M ⊗ P )⊕ (N ⊗ P ), so f j is the identity, and f and j are inverse isomorphisms.

iii*) For any A-modules Mi (i ∈ I ) and N,

N ⊗⊕

i∈I

Mi∼=⊕

i∈I

(N ⊗Mi ).1

Proof. For each finite subset J ⊆ I , write MJ :=⊕

j∈J M j . Then MJ +MJ ′ = MJ∪J ′ , for all finite J , J ′ ⊆ I . Taking allmaps to be the natural insertions, [2.17] shows M ∼= lim−→ MJ . Similarly, for J ⊆ J ′, the natural injection

⊕

j∈J (N ⊗

M j )→⊕

j∈J ′(N ⊗M j ), can be viewed as an inclusion, and [2.17] again says lim−→J

⊕

j∈J (N ⊗M j )

∼=⊕

i∈I (N ⊗Mi ).

Thus

N ⊗⊕

i∈I

Mi

[2.17]∼= N ⊗ lim−→J

MJ

[2.20]∼= lim−→J

(N ⊗MJ )(2.14.iii)∼= lim−→

J

⊕

j∈J

(N ⊗M j ) [2.17]∼=

⊕

i∈I

(N ⊗Mi ).

1 This generalizes (2.14.iii) and uses independent later exercises regarding direct limits.

26

Chapter 2: Modules

iv) A⊗M →M , a⊗ x 7→ ax.

By the definition of an A-module M there is a bi-additive map µ : A× M → M , A-linear in the first variable.The required identity µ

a, µ(b , m)

= µ(ab , m) shows that µ is A-linear in the second variable if we consider Mto have the A-module structure induced by µ. Thus, with g : A×M → A⊗M the canonical map, we get a uniquefactorization µ = µ′ g , where µ′ : A⊗M → M is A-linear and µ′(a⊗ x) = ax. On the other hand there is also anobviously A-linear map ι : M →A⊗M given by x 7→ 1⊗ x. We check that these maps are inverse:

ι

µ′(a⊗ x)

= ι(ax) = 1⊗ ax = a⊗ x and µ′

ι(x)

=µ′(1⊗ x) = x.

Exercise 2.15. Let A, B be rings, let M be an A-module, P a B-module, and N an (A, B)-bimodule (that is, N is simulta-neously an A-module and a B-module and the two structures are compatible in the sense that a(xb ) = (ax)b for all a ∈A,b ∈ B, x ∈N). Then M ⊗A N is naturally a B-module, N ⊗B P an A-module, and we have

(M ⊗A N )⊗B P ∼=M ⊗A (N ⊗B P ).

As in the proof of (2.14.iii), for each z ∈ P we have a map fz : M×N →M⊗A (N⊗B P ) given by ⟨x, y⟩ 7→ x⊗(y⊗z),which we claim is A-bilinear in the first two variables. Bi-additivity is clear, and for a ∈A we have

fz (ax, y) = ax ⊗ (y ⊗ z) = a

x ⊗ (y ⊗ z)

= afz (x, y),

fz (x, ay) = x ⊗ (ay ⊗ z) = x ⊗ a(y ⊗ z) = a

x ⊗ (y ⊗ z)

= afz (x, y).

Thus each fz induces an A-linear map fz : M ⊗A N →M ⊗A (N ⊗B P ) taking x⊗ y 7→ x⊗ (y⊗ z). Allowing z to vary,we have a bi-additive map g : (M ⊗A N )× P → M ⊗A (N ⊗B P ) taking ⟨x ⊗ y, z⟩ 7→ fz (x ⊗ y). This g is obviouslyA-linear in the first variable, and is B -bilinear since for b ∈ B we have

g

(x ⊗ y)b , z

= g (x ⊗ yb , z) = x ⊗ (yb ⊗ z) = x ⊗ (y ⊗ z)b =

x ⊗ (y ⊗ z)

b = g (x ⊗ y, z)b ,

g (x ⊗ y, zb ) = x ⊗ (y ⊗ zb ) = x ⊗ (y ⊗ z)b =

x ⊗ (y ⊗ z)

b = g (x ⊗ y, z)b .

Thus, by the universal property, g gives rise to an (A, B)-linear map g : (M ⊗A N )⊗B P → M ⊗A (N ⊗B P ) taking(x ⊗ y)⊗ z 7→ x ⊗ (y ⊗ z). A symmetric argument gives the inverse map x ⊗ (y ⊗ z) 7→ (x ⊗ y)⊗ z.

Exercise 2.20. If f : A→ B is a ring homomorphism and M is a flat A-module, then MB = B ⊗A M is a flat B-module.

Let j : N1 N2 be any injective B -module homomorphism. By (2.19), to show MB is a flat B -module it sufficesto show j ⊗ idMB

: N1⊗B MB → N2⊗B MB is injective. By restricting scalars along f , we can consider all modules asA-modules, and find canonical A-module isomorphisms

Ni ⊗B MB =Ni ⊗B (B ⊗A M )(2.15)∼= (Ni ⊗B B)⊗A M

(2.14.i)∼=(2.14.iv)

Ni ⊗A M.

Since j is still injective considered as an A-module homomorphism and M is flat, j ⊗ idM : N1⊗A M → N2⊗A M isinjective. Composing on both sides with the canonical isomorphisms yields

x ⊗ y 7→ (x ⊗ 1)⊗ y 7→ x ⊗ (1⊗ y) 7→ j (x)⊗ (1⊗ y) 7→

j (x)⊗ 1

⊗ y 7→ j (x)⊗ y

which then must also be injective; but this is j ⊗ idMB.

Proposition 2.21*. The direct sum M of a family of A-modules Mi (i ∈ I ) is characterized up to isomorphism by thefollowing universal property: there exists a family of homomorphisms ji : Mi → M such that for any A-module N andfamily of homomorphisms fi : Mi →N, there exists a unique homomorphism f : M →N such that fi = f ji for all i ∈ I .

For each x = ⟨xi ⟩ ∈ M , write pi (x) = xi ∈ Mi ; then the projections pi : M → Mi are surjective homomorphisms.For each xi ∈ Mi , write ji (xi ) for the unique element y ∈ M such that pi (y) = xi and pt (y) = 0 for all t 6= i . Theseinsertions ji : Mi →M are injective homomorphisms. Note that pi ji = idMi

, while pi jt = 0 for t 6= i .Since for any x ∈ M we have only finitely many pi (x) 6= 0, it follows x =

∑

i∈I ji

pi (x)

, or idM =∑

( ji pi ).Thus for any homomorphism f : M →N we have

f = f idM = f

∑

( ji pi )

=∑

( f ji pi ).2

2 So f : ⟨xi ⟩ 7→∑

fi (xi ).

27

Ex. 2.1 Chapter 2: Modules

Thus f is uniquely determined by the maps f ji pi , and since each pi : M → Mi is surjective, by the maps f ji .On the other hand, given an arbitrary family of maps fi : Mi →N, we can define f : M →N by f =

∑

t ( ft pt ), andprecomposing with ji , we get f ji =

∑

t ( ft pt ji ) = fi pi ji = fi . Thus M satisfies the universal property.Now suppose another module M ′ and family of homomorphisms j ′i : Mi M ′ also have this property. Then

associated to the maps ji : Mi → M , there is a unique u : M ′ → M such that each ji = u j ′i , and associated to themaps j ′i : Mi →M ′, there is a unique u ′ : M →M ′ such that each j ′i = u ′ ji . It follows each (u u ′) ji = u j ′i = ji .By assumption, associated to the ji : Mi → M there is a unique map j : M → M such that ji = j ji ; since both idMand u u ′ meet this criterion, it follows that the two are equal. Symmetrically, u ′ u = idM ′ , so M ∼=M ′.

Proposition 2.22*. Let fi : Mi → Ni (i ∈ I ) be a family A-module homomorphisms, and M and N the respective directsums of the Mi and the Ni . Write ji for the insertions Mi M , ki for the insertions Ni N, pi for the projectionsM Mi , and qi for the projections N Ni . Then there is a unique direct sum map f =

⊕

i∈I fi : M →N such that

fi = qi f ji and qi f jt = 0 for i 6= t .

Moreover,i) f is injective if and only if each fi is injective;ii) f is surjective if and only if each fi is surjective.

The map ⟨xi ⟩ 7→

fi (xi )

satisfies the conditions on f .3 Since any other g satisfying the equations takes ji (xi ) toki

fi (xi )

, it follows by additivity that g : ⟨xi ⟩ 7→

fi (xi )

for all ⟨xi ⟩ ∈M , so g = f .4

i): Suppose all fi are injective. If f

⟨xi ⟩

=

fi (xi )

= 0, then each fi (xi ) = 0, so each xi = 0, and ⟨xi ⟩= 0.5

If some fi takes a nonzero xi to 0, then f takes its image j (xi ) to 0, and so is also not injective.6

ii): Let y = ⟨yi ⟩ ∈ N be given. By assumption, there is for each i an xi ∈ Mi with fi (xi ) = yi ; if yi = 0, we maytake xi = 0. Then x = ⟨xi ⟩ ∈M and f (x) = y, so f is surjective.7

If yi ∈Ni is not in im fi , then ki (yi ) cannot be in the image of f : ⟨xi ⟩ 7→

fi (xi )

, so f is not surjective.8

EXERCISES1. Show that (Z/mZ)⊗Z(Z/nZ) = 0 if m, n are coprime.

Since m, n are coprime, by Bézout’s lemma there are a, b ∈Z such that am+bn = 1.9 Let x⊗y ∈ (Z/mZ)⊗Z(Z/nZ).Then x ⊗ y = (am+ bn)(x ⊗ y) = a(mx ⊗ y)+ b (x ⊗ ny) = 0.

2. Let A be a ring, a an ideal, M an A-module. Show that (A/a)⊗A M is isomorphic to M/aM.

Applying the right exact functor−⊗A M to the short exact sequence 0→ a→A→A/a→ 0, we see the sequence

a⊗Mj→ A⊗M → (A/a)⊗M → 0

is exact, so(A/a)⊗M ∼= (A⊗M )/im j.

But the absorption isomorphism A⊗M →M of (2.14.iv) sends im j → aM, so (A⊗M )/im j ∼=M/aM.

3 More formally, by (2.21), there is a unique f : M → N with f ji = ki fi : Mi → N. Now each qi f ji = qi ki fi = fi , andqi f jt = (qi kt ) ft = 0 for t 6= i .

4 Suppose that g also satisfies the first equation. Then each ki fi = ki qi g ji = g ji , so by uniqueness in (2.21), f = g .5 Alternately, f =

∑

i , t ki qi f jt pt =∑

ki qi f ji pi =∑

ki fi pi . Since (im qi ) ∩ (∑

t 6=i im qt ) = 0, we know ker f =⋂

ker(ki fi pi ). Since each ki fi is injective, this is⋂

ker pi = 0.6 If f =

∑

kt ft pt is injective, so is f ji =∑

t kt ft pt ji = ki fi . Since ki is injective, so is fi .7 Proceeding formally, since each fi = qi f ji is surjective, so is each qi f . Since idN =

∑

ki qi , we have f = idN f =∑

ki qi f , soim f =

∑

im(ki qi f ) =∑

ki (Ni ) =N.8 Since fi pi = qi f ji pi = qi f

∑

( jt pt ) = qi f , if some yi ∈ Ni is not in im fi , it is not in the image of fi pi = qi f , and soki (yi ) ∈Ni is not in the image of ki qi f = f .

9 [MWBezout]; Another way of putting this is that m and n being coprime in the arithmetic sense of having no common irreducible factorsimplies that (m) and (n) are coprime in the algebraic sense (p. 7) that (m)+ (n) = (1).

28

Chapter 2: Modules Ex. 2.3

3. Let A be a local ring, M and N finitely generated A-modules. Prove that if M ⊗N = 0, then M = 0 or N = 0.

Let m be the maximal ideal and k = A/m the residue field. The scalar extensions Mk := k ⊗A M and Nk arek-vector spaces. That M ⊗N = 0 implies

Mk ⊗Nk = (k ⊗M )⊗ (k ⊗N )(2.14.iii)∼=(2.14.iv)

k ⊗ (M ⊗N ) = (M ⊗N )k = 0.

But dimension of vector spaces is multiplicative under tensor, so Mk or Nk = 0. Without loss of generality, assumeMk = 0. By [2.2], Mk

∼= M/mM , so mM = M . By Nakayama’s Lemma (2.6), since M is finitely generated and m isthe Jacobson radical, we have M = 0.

4. Let Mi (i ∈ I ) be any family of A-modules, and let M be their direct sum. Prove that M is flat ⇐⇒ each Mi is flat.

Let an A-linear map j : N ′ → N be given. Using the isomorphisms of (2.14.iii*) identifies j ⊗ idM with a maph :⊕

i∈I (N′⊗Mi )→

⊕

i∈I (N′⊗Mi ), and the compositions

N ′⊗Mi ⊕

i∈I

(N ′⊗Mi )∼−→N ′⊗M

j⊗idM−−→N ⊗M∼−→⊕

i∈I

(N ⊗Mi )N ⊗Mi

are j ⊗ idMi, the associated maps N ′⊗Mi → N ⊗Mt for t 6= i being zero. Thus h is the direct sum of the j ⊗ idMi

,and by (2.22.i*), j ⊗ idM is injective just if they are.

If the Mi are flat and j is injective, then by (2.19) each of the j ⊗ idMiare as well, and so j ⊗ idM is. Hence M is

flat. If Mi is not flat, there exists a j such that j ⊗ idMiis not injective, and so j ⊗ idM is not. Hence M is not flat.

5. Let A[x] be the ring of polynomials in one indeterminate over a ring A. Prove that A[x] is a flat A-algebra.

A is a flat A-module, for by (2.14.iv) the functor −⊗A A is naturally isomorphic to the identity functor. LetMi = Ax i ⊆ A[x] for i ∈ N. Each Mi

∼= A as an A-module, and so is flat. Then as a module, A[x] =⊕

i∈NMi is flatby [2.4].

6. For any A-module M , let M [x] denote the set of all polynomials in x with coefficients in M , that is to say expressions ofthe form

m0+m1x + · · ·+mr x r (mi ∈M ).

Defining the product of an element of A[x] and an element of M [x] in the obvious way, show that M [x] is an A[x]-module.

As an A-module, we have M [x] ∼=⊕

n∈NMxn . We define the action of A[x] on M [x] by (∑

ai xi )(∑

mj xj ) =

∑

ck xk , where ck =∑

i+ j=k ai mj . We check the distributivity and associativity. Let f (x) =∑

i ai x i and g (x) =∑

j b j x j ∈A[x] and m(x) =∑

k mk xk and n(x) =∑

k nk xk ∈M [x]. Associativity is given by

f (x)g (x)

m(x) =

∑

l

∑

i+ j=k

ai b j

xk

∑

l

ml x l

=∑

p

∑

k+l=p

∑

i+ j=k

ai b j

mk

x p =∑

p

∑

i+ j+k=p

ai b j mk

x p ;

f (x)

g (x)m(x)

=

∑

i

ai x i

∑

l

∑

j+k=l

b j mk

x l

=∑

p

∑

i+l=p

ai

∑

j+k=l

b j mk

x p =∑

p

∑

i+ j+k=p

ai b j mk

x p .

Distributivity is given by

f (x)+ g (x)

m(x) =

∑

i

(ai + bi )xi

∑

k

mk xk

=∑

l

∑

i+k=l

(ai mk + bi mk )

x l

=

∑

i

ai x i

∑

k

mk xk

+

∑

i

bi x i

∑

k

mk xk

= f (x)m(x)+ g (x)m(x);

f (x)

m(x)+ n(x)

=

∑

i

ai x i

∑

k

(mk + nk )xk

=∑

l

∑

i+k=l

(ai mk + ai nk )

x l

=

∑

i

ai x i

∑

k

mk xk

+

∑

i

ai x i

∑

k

nk xk

= f (x)m(x)+ f (x)n(x).

29


Show M [x]∼=A[x]⊗AM .

Define φ : M [x]→ A[x]⊗A M by m(x) =∑

m j x j 7→∑

(x j ⊗m j ). It is obviously additive, and is A[x]-linear,for if f (x) =

∑

ai x i ∈A[x], then

φ

f (x)m(x)

=∑

k

∑

i+ j=k

φ(ai mj xk ) =∑

k

∑

i+ j=k

(xk ⊗ ai mj ) =∑

i

∑

j

(x i x j ⊗ ai m j )

=∑

j

∑

i

ai x i

x j ⊗mj

=

∑

i

ai x i

∑

j

x j ⊗mj

= f (x)φ

m(x)

.

Define ψ : A[x]×M → M [x] by ψ∑

ai x i , m

=∑

(ai m)xi . It is clearly bi-additive and A-bilinear, and so induces

a linear map ψ : A[x]⊗A M →M [x] sending∑

ai x i

⊗m 7→∑

(ai m)xi . Now φ and ψ are inverse, for

ψ

φ(mi x i )

=ψ(x i ⊗mi ) = mi x i

andφ

ψ(ai x i ⊗m)

=φ

(ai m)xi = x i ⊗ ai m = ai x i ⊗m.

7. Let p be a prime ideal in A. Show that p[x] is a prime ideal in A[x]. If m is a maximal ideal in A, is m[x] a maximalideal in A[x]?

p[x] is the kernel of the “reduction of coefficients” homomorphism A[x] (A/p)[x], and (A/p)[x] is an integraldomain (see the proof of [1.2.ii]).

On the other hand, the ideal (2) Ã Z is maximal, but the ideal 2Z[x] Ã Z[x] is not maximal, as the quotient(Z/2Z)[x] is not a field. 2Z[x] is properly contained in the maximal ideal (2, x).

8. i) If M and N are flat A-modules, then so is M ⊗AN.

Let j : P ′ P be an injective A-linear map. Since N is flat, the map idN ⊗ j : N ⊗ P ′→ N ⊗ P is injective. SinceM is flat, the map idM ⊗ (idN ⊗ j ) : M ⊗ (N ⊗ P ′)→M ⊗ (N ⊗ P ) is injective. But by the associativity (2.14.ii) of ⊗A,this is up to a canonical isomorphism the map idM⊗N ⊗ j induced from j by tensoring with M ⊗N , so M ⊗N is flat.

ii) If B is a flat A-algebra and N is a flat B-module, then N is flat as an A-module.

Let j : M ′ M be an injective A-module homomorphism, and let f : A→ B be the map making B an A-algebra.Since B is a flat A-module, the map idB ⊗A j : B ⊗A M ′ → B ⊗A M is injective, and since N is flat as a B -module, themap

idN ⊗B (idB ⊗A j ) : N ⊗B (B ⊗A M ′)→N ⊗B (B ⊗A M )

is injective as well. Composing the associativity isomorphisms of (2.15), we see

(idN ⊗B idB )⊗A j : (N ⊗B B)⊗A M ′→ (N ⊗B B)⊗A M

is injective, so by the isomorphism N ∼=N ⊗B B of (2.14.iv), so is idN ⊗A j : N ⊗A M ′→N ⊗A M .

9. Let 0→M ′→M →M ′′→ 0 be an exact sequence of A-modules. If M ′ and M ′′ are finitely generated, then so is M .

Without loss of generality view M ′→ M as an inclusion and M → M ′′ as a quotient mapping. Let the finite setsxii and y j j respectively generate M ′ and M ′′. Lift the y j to elements y j of M . The submodule of M generated bythe finite set xii ∪y j j contains M ′ and has image M ′′, so by the bijection (p. 18) between submodules of M ′′ andsubmodules of M containing M ′, it is M .

10. Let A be a ring, a an ideal contained in the Jacobson radical of A; let M be an A-module and N a finitely generated A-module, and let u: M →N be a homomorphism. If the induced homomorphism M/aM →N/aN is surjective, then u issurjective.

As the induced homomorphism sends M M/aM N/aN, we must have u(M )+aN =N. Since N is finitelygenerated and a⊆R, by the corollary (2.7) of Nakayama’s Lemma, u(M ) =N.

30


11. Let A be a ring 6= 0. Show that Am ∼=An =⇒ m = n.

Letφ : Am→An be an isomorphism andmÃAa maximal ideal. If k =A/m is the quotient field, then idk⊗φ : k⊗AAm → k ⊗AAn is an isomorphism by (2.18), taking N = k, M ′ = 0, M = Am , and M ′′ = An . But by (2.8), we havek⊗AAn ∼= kn an n-dimensional k-vector space. Since dimension of vector spaces is an isomorphism invariant, m = n.

If φ : Am→An is surjective, then m ≥ n.

As above, tensoring with k =A/m shows that the k-linear map idk⊗φ : k m→ kn is surjective. But if m < n, them elements φ(ei ) cannot span kn , so m ≥ n.

If φ : Am→An is injective, is it always the case that m ≤ n?

It is indeed the case. Some poached solutions follow.10

i)11 This solution is simplest and uses results already proven in the book by Ch. 2. Letφ : Am→An be an A-linearmap with m > n; we prove it is not injective. Compose with the inclusion i : An =An×0m−n ,→An×Am−n =Am

on the first n coordinates to get an A-module endomorphism ψ= i φ : Am → Am . If π : Am → A is the projectionon the last coordinate, we have π ψ= 0. Now by (2.4), ψ satisfies an equation

ψn + a1ψn−1+ · · ·+ an idAm = 0

for some a j ∈A. Assume n is minimal such this happens. Takingπ of both sides, we see an = 0. Now asψ is A-linear,we haveψ(ψn−1+a1ψ

n−2+ · · ·+an−1 idAm ) = 0. Since n was minimal, the mapψn−1+a1ψn−2+ · · ·+an−1 idAm 6= 0,

so its image M is not 0, yet ψ(M ) = 0, so ψ (and hence φ) is not injective.

ii)12 The other proof feasibly accessible using knowledge available so far uses some linear algebra, generalized tothe context of free modules over a commutative ring A. Given a square matrix N of rank n with entries ai j ∈ A,the determinant detA is the element of A given by

∑

σ (sgnσ)∏n

i=1 ai ,σ(i) where σ ranges over all n! permutationsof 1, . . . , n and sgnσ is the parity of the permutation, which is ±1 depending as σ is even or odd. From thisformula it follows that if two rows of N are identical, the determinant is 0. Note that there are n2 square matricesNi j of rank n − 1 given by deleting the entries in the i th row and j th column. The (i , j )-cofactor of N is given byci j = (−1)i+ j detNi j ∈ A. The determinant of N can be calculated recursively by the cofactor expansion detN =∑n

i=1 ai j ci j for fixed j or∑n

j=1 ai j ci j for fixed i . The adjugate Adj(N ) = (bi j ) of N is the n× n matrix with entriesbi j = c j i the cofactors of N . The (i , j ) entry of N ·Adj(N ) is

∑nk=1 ai k ck j =

∑

k ai k b j k . For i = j , the cofactorexpansion of the determinant shows this number is detN . For i 6= j this expression for the entry is, up to a sign, thecofactor expansion, along the i th row, of the determinant of the matrix

a11 · · · a1n...

. . ....

a j 1 · · · a j n...

. . ....

a j 1 · · · a j n...

. . ....

an1 · · · ann

,

whose i thand j th rows are equal; and so the entry is 0. Thus N ·Adj(N ) = det(N ) · In is the scalar product of detNand the n× n identity matrix.

Note that it suffices to prove an A-module homomorphismφ : Am→An cannot be injective for m = n+1, sinceif n ≤ m− 1 we could compose with the inclusion An ,→ Am−1 to transform an injection Am An to an injectionAm Am−1. If ei (i = 1, . . . , n+ 1) is the standard basis for An+1 and f j ( j = 1, . . . , n) is the standard basis for An ,

10 Several solutions are up at http://mathoverflow.net/questions/136/atiyah-macdonald-exercise-2-11/2622. I was unable tofind a solution myself, at least before giving up and searching online.

11 Balazs Strenner12 Robin Chapman

31

http://mathoverflow.net/questions/136/atiyah-macdonald-exercise-2-11/2622


then φ(ei ) =∑n

j=1 a j i f j for some a j i ∈A, and φ is represented by the matrix

M =

a11 · · · a1, n+1...

. . ....

an, 1 · · · an, n+1

,

so φ is injective just if there is no nonzero vector v = (b1, . . . , bn+1)> ∈ An+1 such that Mv = 0. Now let Mi be the

n× n matrix obtained from M by deleting the i th column and let v have components bi = (−1)i det Mi . Then thej th component of Mv is

∑n+1i=1 (−1)i a j i det Mi . But this is (−1) j times the cofactor expansion along the j th row of the

determinant of the (n+ 1)× (n+ 1)matrix

a11 · · · a1, n+1...

. . ....

a j 1 · · · a j , n+1a j 1 · · · a j , n+1...

. . ....

an1 · · · an, n+1

,

which is zero because the j th row is repeated, so Mv = 0.Now if some det Mi is nonzero, we have achieved our goal of finding a nonzero v ∈ kerφ. Otherwise, det Mn+1 =

0. If Mn+1 has a nonzero vector v ′ = (b1, . . . , bn)> in its kernel, then v = (b1, . . . , bn , 0)> is a nonzero vector in the

kernel of M . It then falls to us to show that if a square matrix N of rank n has determinant 0, it has nontrivial kernel.Let r < n be the rank of the largest square submatrix (obtained from N by deleting rows and columns) with nonzerodeterminant; by shuffling rows, we may assume that r× r occurs in the upper left of N . Let R be the (r+1)×(r+1)matrix on the upper left of N containing it; since det R 6= 0 by the maximality of r , taking the cofactor expansionalong the first column of R shows that the first column v ′′ of Adj(R) has some nonzero entry. Now Rv ′′ = 0, sinceR ·Adj(R) = det(R) · Ir+1 = 0. If we let v ′ be v ′′ with n− r zeros added at the end, then as the determinant of N iszero, the rest of the rows of N are linear combinations of rows of R, so N v ′ = 0.

iii)13 Abstracting from the last proof at a rather high level is the following. It requires the notion of exteriorproduct:

∧nA M =

⊗n

A M

/N , where⊗n

A M is the n-fold tensor product M ⊗A · · · ⊗AM and N is the submodulegenerated by all elements (· · · ⊗ x ⊗ y ⊗ · · · ) + (· · · ⊗ y ⊗ x ⊗ · · · ). The image of x ⊗ · · · ⊗ y is denoted by x ∧ · · · ∧ y,and we have (by fiat) the equalities x1 ∧ · · · ∧ xn = (sgnσ)xσ(1) ∧ · · · ∧ xσ(n), where σ is a permutation of 1, . . . , nand sgnσ its parity. We then have a theorem:14

a subset u1, . . . , um of M =An is linearly independent ⇐⇒ ∀a ∈A [a · (u1 ∧ · · · ∧ um) = 0 =⇒ a = 0],

where u1 ∧ · · · ∧ um ∈∧n

A M . This proves the result because for m > n we have∧m

A (An) = 0.

The remaining proofs use material developed later in the book.iv)15 Letφ : AmAn be an injective A-module homomorphism represented by the matrix M . Let B =Z[ . . . , ai j , . . .]

be the subring of A generated by all the entries of the matrix M ; since Z is Noetherian, by (7.5) (the Hilbert BasisTheorem) and (7.1) (quotient preserves a.c.c.), B is a Noetherian ring; and φ restricts to an injective linear mapψ : B m B n . Note that B n , by (6.4), is Noetherian. If we assume m > n we can derive a contradiction. WriteB m = M ⊕ N with M ∼= B n and N ∼= B m−n . Then we have isomorphic images M1 and N1 of M , N in M , andisomorphic images M2, N2 in M1, and isomorphic images M3, N3 in M2, etc. This yields an infinite ascending chain

N1 (N1⊕N2 (N1⊕N2⊕N3 ( · · · ,contradicting the ascending chain condition.

v)16 Continue with the Noetherian ring B of iv). By [1.8], B has a minimal prime ideal p. By p. 38 the localizationC = Bp has only one prime ideal q= pC . By (3.3) (localization is exact), the induced map ψp : C m→C n is injective

13 Pete L. Clark, via Tsit Yuen Lam’s Lectures on Rings and Modules, pp. 15–16, via Nicolas Bourbaki’s Algebra14 Nicolas Bourbaki, Algebra, Chapter III, §7.9, Prop. 12, page 51915 from Tsit Yuen Lam’s Lectures on Rings and Modules, p. 14, and referred by Pete L. Clark16 Georges Elencwajg

32


as well. By (7.3) (localization preserves a.c.c.), C is Noetherian, and since it has just one prime ideal, it has (Krull)dimension (p. 90) zero. By (8.5), C is Artinian as well, so by (6.8) it has a finite composition series as a C -module. By(6.7), the length of this composition series is independent of the series chosen, so C has a well defined finite lengthl (C )≥ 1. Now let us consider the lengths of C m and C n . (6.9) says the length of a module is an additive function, sousing the natural exact sequences 0→C →C n+1→C n → 0, we have l (C n) = n · l (C ) by induction. We also have,by assumption, an exact sequence

0→C m ψp−→C n→ coker(ψp)→ 0,

so n · l (C ) = m · l (C )+ l

coker(ψp)

, and thus m ≤ n.vi)17 Finally, there is another matrix-theoretic proof, involving localization.φ : An+1→An is injective just if each

of its localizations is injective, by (3.9). By [1.8], A has a minimal prime ideal p, and by [1.10], q= pAp is the nilradicalin the localization B = Ap, and all other elements of B are units. We claim any finite set of elements in q is jointlyannihilated by some nonzero element of q. For the base case, if 0 6= x ∈ q and r > 0 is minimal such that x r = 0,then x r−1 ∈ q is a nonzero element annihilating x. Let 0 6= y ∈Ann(S) for a finite set S ⊆ q, and let z ∈ q\0. Thereis r > 0 such that z r = 0, and so there is a minimal s ∈ [1, r ] such that y z s = 0. Then 0 6= y z s−1 ∈ Ann

S ∪ z

.This lemma essentially allows us to use Gaussian elimination.

Let M = [a j i ] be the matrix of the B -module homomorphism ψ= φp : B n+1→ B n . The columns represent theimagesψ(ei ), which we are linearly independent just ifψ is injective. Now if there is some linear dependency amongthe columns, then adding a multiple of one column to another preserves the existence of the dependency. Sincethe operation of adding a multiple of one column to another is invertible (subtract a multiple of the column), thiscolumn operation also reflects dependence, so the columns of the altered matrix are independent just if the columnsof the original are. Clearly the same holds for the operations of multiplying all the entries of a column by a unit,swapping rows, and swapping columns.

If any column contains no unit, then by the claim above, there is a nonzero a annihilating that column, and thevector (0 · · · a · · · 0)> is killed by M , contradicting injectivity of ψ. Thus if ψ is injective all columns contain someunit. If the first column contains a unit, we may shuffle rows so that a11 is a unit, and multiplying the first columnby a−1

11 we may assume a11 = 1. Subtracting multiples of this first column from the others we may clear the rest ofthe first row. If the second column contains a unit, by swapping rows we may assume it is a22. Multiplying by a unit,we can assume a22 = 1, and subtracting multiples of the second column from the other columns, we may clear therest of the second row.

Carrying on in this fashion, we either come upon a column containing no unit or transform the matrix to theform

1 0 · · · 0 00 1 · · · 0 0...

.... . .

......

0 0 · · · 1 0

,

contradicting injectivity.Note there exist abelian groups G isomorphic to Gn for n = 1, 2, . . . , ℵ0. Let G =

∏

j∈NH j be the direct prod-uct of infinitely many copies H j of some abelian group H . Since N is infinite, there are for each n ∈ N bijectionsφn : 1, . . . , n ×N↔ N, and there is a bijection φω : N×N↔ N. These yield, for n ∈ N∪ ℵ0, group isomor-phisms ψn : Gn ∼−→ G by letting ψ(x)φn (i , j ) = (xi ) j , where xi ∈ G =

∏

NH , so (xi ) j ∈ H . This shows that if weweaken the definition of a ring to that of a “rng” (“ring without identity”: ⟨A, ·⟩ is only required to be a semigroup,not a monoid), the proposition doesn’t hold. For it is possible to have a nonzero rng whose product is identicallyzero: any additive abelian group G gives rise to a rng with trivial multiplication. It is then legitimate to define aG-module structure on another abelian group M by g ·m = 0.

12. Let M be a finitely generated A-module and φ : M → An a surjective homomorphism. Show that ker(φ) is finitely gener-ated.

Lemma.* Let A be a ring and M and N be A-modules. If there exist homomorphisms s : N →M and r : M →N suchthat r s = idN , then M ∼=N ⊕ (coker s)∼=N ⊕ (ker r).

Define the map κ : M →N ⊕ (coker s) by x 7→

r (x), x

. For injectivity, suppose x ∈ kerκ. Then x = 0, so thereis y ∈ N such that x = s(y), and 0 = r (x) = rs(y) = y, so x = 0. For surjectivity, let y ∈ N and x ∈ coker s be

17 Karl Dahlke, http://mathreference.com/mod-pit,basec.html#embed

33

http://mathreference.com/mod-pit,basec.html#embed


arbitrary, and let x ∈M be some lift of x. If z = x + s

y − r (x)

, then z = x, while r (z) = r (x)+ rs(y)− rsr (x) =r (x)+ y − r (x) = y.

An isomorphism M →N ⊕ker r is given by λ= ⟨r, idM −sr ⟩. Indeed, r (idM −sr ) = r − rsr = 0; and x ∈ kerλimplies r (x) = 0 and x = x − sr (x) = 0; and for any y ∈ N and z ∈ ker r , if we let w = s(y) + z, then λ(w) =

rs(y)+ r (z), s(y)+ z − sr s (y)− sr (z)

= ⟨y, s(y)+ z − s(y)⟩= ⟨y, z⟩.

Now we will show kerφ is a quotient (actually a summand) of M ; hence the images of a finite set of generatorsfor M will generate kerφ. Let e1, . . . , en be a basis for An and pick any elements χ (ei ) ∈ φ−1(ei ); this extends to ahomomorphism χ : An → M such that φχ = idAn . Define ψ = idM −χφ: it takes M → kerφ since φχφ = φ. Forx ∈ kerφ we have ψ(x) = x, so ψ is surjective. In fact, writing ι : kerφ ,→ M for the inclusion, φ ι = idkerφ, sokerφ is a summand, with coker ι=M/kerφ∼= imφ=An as the other summand.

13. Let f : A→ B be a ring homomorphism, and let N be a B-module. Regarding N as an A-module by restriction of scalars,form the B-module NB = B⊗AN. Show that the homomorphism g : N →NB which maps y to 1⊗ y is injective and thatg (N ) is a direct summand of NB .

The quotient map NB = B ⊗AN B ⊗B N is also a B -module homomorphism, since b (b ′ ⊗ y) = bb ′ ⊗ y 7→bb ′⊗ y = b (b ′⊗ y), and composing with the isomorphism (2.14.iv*) gives a B -module homomorphism h : NB →Ntaking b ⊗ y 7→ by. Now hg = idN , so by the lemma in (2.12*), g injects N as a summand of NB .

Direct limits14. A partially ordered set I is said to be a directed set if for each pair i , j in I there exists k ∈ I such that i ≤ k and j ≤ k.

Let A be a ring, let I be a directed set and let (Mi )i∈I be a family of A-modules indexed by I . For each pair i , j in Isuch that i ≤ j , let µi j : Mi →M j be an A-homomorphism, and suppose that the following axioms are satisfied:

(1) µi i is the identity mapping of Mi , for all i ∈ I ;(2) µi k =µ j k µi j whenever i ≤ j ≤ k.

Then the modules Mi and homomorphisms µi j are said to form a direct system M= (Mi , µi j ) over the directed set I .We shall construct an A-module M called the direct limit of the direct system M. Let C be the direct sum of the Mi ,

and identify each module Mi with its canonical image in C . Let D be the submodule of C generated by all elements of theform xi −µi j (xi ) where i ≤ j and xi ∈Mi . Let M =C/D, let µ : C →M be the projection, and let µi be the restrictionof µ to Mi .

The module M , or more correctly the pair consisting of M and the family of homomorphismsµi : Mi →M , is called thedirect limit of the direct system M, and is written lim−→Mi . From the construction it is clear that µi =µ j µi j wheneveri ≤ j .

In case it wasn’t clear, let i ≤ j and xi ∈ Mi : then xi −µi j (xi ) ∈ D = ker(µ), so µi (xi ) = µ(xi ) = µ

µi j (xi )

=µ j (µi j (xi )).

15. In the situation of Exercise 14, show that every element of M can be written in the form µi (xi ) for some i ∈ I and somexi ∈Mi .

If (I , ≤) is directed and S ⊆ I is finite, then by induction there is a j ∈ I such that for each i ∈ S we have i ≤ j .Surely this is the case if j = i ∈ i = S, and if it is the case for a given S and we add a new element in+1 to S, thenthere is an element k ≥ in+1, j , and so k ≥ every element of S ∪in+1.

Let x ∈ M ; then it is the image under the quotient map C → C/D = M of some sum∑

i∈S xi ∈ C =⊕

i∈I Mi ,where S ⊆ I is finite by definition. Pick a j ∈ I such that for all i ∈ S we have j ≥ i . The elements µi j (xi )− xi ∈D ,so ex j =

∑

i∈S µi j (xi )≡∑

i∈S xi (mod D) and µ j (ex j ) =µ(ex j ) = x.

Show also that if µi (xi ) = 0 then there exists j ≥ i such that µi j (xi ) = 0 in M j .We first assemble some auxiliary information about the module D . First, given any generator (id−µi j )(xi ) of

D , multiplying by a ∈ A we get (id−µi j )(axi ), since the µi j are A-module homomorphisms. Similarly, given gen-erators (id−µi j )(xi ) and (id−µi j )(yi ), their sum is a generator (id−µi j )(xi + yi ). Thus, in considering expressions∑n

k=1 ak

xik−µik jk

(xik)

it suffices to assume each ak = 1 and each pair (ik , jk ) only occurs once.Now suppose that xi ∈Mi is such that µi (xi ) =µ(xi ) = 0. Then xi ∈Mi ∩D , so xi can be written as a finite sum

xi =∑

( j ,k)∈T

y j −µ j k (yk )

=∑

j∈S

z j (2.5)

34


for some finite set S ⊆ I , some set T ⊆ S2 of pairs ( j , k)with j ≤ k, and some elements y j , z j ∈M j . Here the middlesum expresses xi ∈ D in terms of generators for D , and the right sum breaks the middle sum into components inM j . Since Eq. 2.5 takes place in the direct sum, we must have cancellation in all components but the i th, so zi = xi ,and for j 6= i we have z j = 0. Let ` ∈ I be an element such that ` ≥ k for each k ∈ S. Then using the equationsµ j` =µk` µ j k ,

µi`(xi ) =∑

j∈S

µ j`(z j ) =∑

( j , k)∈T

µ j`(y j )−µk`

µ j k (y j )

= 0.

16. Show that the direct limit is characterized (up to isomorphism) by the following property. Let N be an A-module and foreach i ∈ I let αi : Mi → N be an A-module homomorphism such that αi = α j µi j whenever i ≤ j . Then there exists aunique homomorphism α : M →N such that αi = α µi for all i ∈ I .

I think we should add the requirement on (M , µi , µi j ) that we have µi =µ j µi j for all i ≤ j ∈ I .First we show that M = lim−→Mi satisfies this property. We showed µi = µ j µi j in [2.14]. Given arbitrary

A-module homomorphisms αi : Mi → N , we have by the universal property of direct sums a unique induced ho-momorphism eα : C =

⊕

i∈I Mi → N . For any xi ∈ Mi and j ≥ i , consider the generator xi −µi j (xi ) of D . By thedefinition of eα and the compatibility condition on the αi we have

eα

xi −µi j (xi )

= αi (xi )−α j

µi j (xi )

= αi (xi )−αi (xi ) = 0,

so D ⊆ ker(eα) and eα induces an A-module homomorphism α : M =C/D→N . Moreover, by definition α

µi (xi )

=eα(xi ) = αi (xi ).

Mi

µi j

µi=βi

µ′i=αi

##M

γ α // M ′

βoo

M j

µ j=β j

??

µ′j=α j

<<

Now suppose that (M , µi : Mi →M ) and (M ′, µ′i : Mi →M ′) both satisfy the univer-sal mapping property. Since (M ′, µ′i ) is a direct limit of (Mi , µi j ) we have by definitionthat µ′i = µ

′j µi j . But then setting αi = µ

′i in the universal property of (M , µi ) as a

direct limit, we get a unique homomorphism α : M → M ′ such that µ′i = αi = α µi .Symmetrically, we get a unique homomorphism β : M ′ → M such that µi = β µ′i .Now µi = β µ′i = β α µi : Mi → M for each i . Since µi = µ j µi j , there exists aunique homomorphism γ : M → M such that µi = γ µi ; as both idM and β α meetthe requirements for γ , by uniqueness, β α = idM . Symmetrically, α β = idM ′ , soα : M ↔M ′ : β are inverse isomorphisms.

17. Let (Mi )i∈I be a family of submodules of an A-module, such that for each pair of indices i , j in I there exists k ∈ I suchthat Mi +M j ⊆Mk . Define i ≤ j to mean Mi ⊆M j and let µi j : Mi →M j be the embedding of Mi in M j . Show that

lim−→Mi =∑

Mi =⋃

Mi .

In particular, any A-module is the direct limit of its finitely generated submodules.For each i we have Mi ⊆

∑

Mi , so⋃

Mi ⊆∑

Mi . On the other hand, if y =∑

i∈I xi ∈∑

Mi is any finite sum,let S = i ∈ I : xi 6= 0, and let j ∈ I be ≥ each element of S; then

∑

i∈S Mi ⊆ M j , so y ∈ M j ⊆⋃

i∈I Mi . Thus∑

Mi =⋃

Mi .We show lim−→Mi

∼=⋃

Mi by showing⋃

Mi has the expected universal property ([2.16]). Let µi : Mi ,→⋃

Mi bethe inclusion, and suppose we have αi : Mi → N such that αi = α j µi j for i ≤ j . This is just the same as sayingthat if Mi ⊆ M j we have α j |Mi

= αi , so that the αi are consistent on all intersections and thus their union defines aunique function α =

⋃

αi :⋃

Mi → N restricting to αi on each Mi ; that is αi = α µi . Now α is A-linear becauseits restriction to each Mi is, so

⋃

Mi satisfies the universal mapping property required of lim−→Mi .It follows that an A-module M is the direct limit of its finitely generated submodules, for any x ∈ M is in the

finitely generated submodule Ax (so M is the union of its finitely generated submodules) and if N1, N2 ⊆ M arefinitely generated submodules, then both are contained in the finitely generated module N1+N2 ⊆M (so the finitelygenerated submodules and inclusions form a direct system).

35


18. Let M= (Mi , µi j ), N= (Ni , νi j ) be direct systems of A-modules over the same directed set. Let M , N be the direct limitsand µi : Mi →M , νi : Ni →N the associated homomorphisms.

M

φ

Mi

αi 44

µi j

//

φi

µi

77

M j

α j

φ j

µ j

>>

Ni

νi j //

νi''

N jν j

N

A homomorphism Φ : M→ N is by definition a family of A-module homomorphismsφi : Mi → Ni such that φ j µi j = νi j φi whenever i ≤ j . Show that Φ defines a uniquehomomorphism φ= lim−→φi : M →N such that φ µi = νi φi for all i ∈ I .

Define αi : Mi →N by αi = νi φi . Then if i ≤ j we have

α j µi j = ν j φ j µi j = ν j νi j φi = νi φi = αi ,

so by the universal property of [2.16] there is a unique mapφ : M →N such thatφµi =αi = νi φi for all i ∈ I .

19. A sequence of direct systems and homomorphisms

M→N→ P

is exact if the corresponding sequence of modules and module homomorphisms is exact for each i ∈ I . Show that thesequence M →N → P of direct limits is then exact.

Let the components of Φ : M→N and Ψ : N→ P be φi : Mi →Ni and ψi : Ni → Pi ,inducing φ : M → N and ψ : N → P , and let the maps in the direct systems M, N, Pbe respectively µi j , νi j , πi j . To show ψ φ = 0, recall ([2.15]) that any element x ∈M is of the form µi (xi ) for some xi ∈ Mi and some i ∈ I . By the assumed exactness,(ψi φi )(xi ) = 0. Then using the defining properties of x j and of φ and ψ ([2.18]),

(ψ φ)(x) = (ψ φ µi )(xi ) = (ψ νi φi )(xi ) = (πi ψi φi )(xi ) =πi (0) = 0.

xi φi //

_µi

ψi //_

νi

0_

πi

x φ // ψ // φ

ψ(x)

Miφi //

µi

Niψi //

νi

Pi

πi

M

φ // N ψ // P

yi ψi //

_νi j

_πi j

x j φ j //

_µ j

νi j (yi ) ψ j //

_ν j

0_π j

x φ // y ψ // 0

Niψi //

νi j

Piπi j

M j

φ j //

µ j

N j

ψ j //

ν j

P jπ j

Mφ // N

ψ // P

On the other hand, suppose y ∈ ker(ψ). By [2.15], there are i ∈ I and yi ∈ Ni suchthat y = νi (yi ), and by the defining property ([2.18]) of ψ we have 0 = ψ(y) = ψ

νi (yi )

=πi

ψi (yi )

. By [2.15] there is j ≥ i such that 0=πi j

ψi (yi )

=ψ j

νi j (yi )

, where we use thedefinition of Ψ being a homomorphism. Since we assumed the sequence of direct systemsis exact, it follows that there is x j ∈M j such thatφ j (x j ) = νi j (yi ). But then if x =µ j (x j )wehave

φ(x) =φ

µ j (x j )

= ν j

φ j (x j )

= ν j

νi j (yi )

= νi (yi ) = y,

using the definitions of x and φ, the assumed property of x j , the result of [2.14] for thedirect system N, and the assumed property of yi . Thus ker(ψ) ⊆ imφ and the sequenceM →N → P is exact.

Tensor products commute with direct limits20. Keeping the same notation as in Exercise 14 let N be any A-module. Then (Mi ⊗ N , µi j ⊗ 1) is a direct system; let

P = lim−→(Mi ⊗N ) be its direct limit. For each i ∈ I we have a homomorphism µi ⊗ 1: Mi ⊗N → M ⊗N, hence byExercise 16 homomorphism ψ : P →M ⊗N. Show that ψ is an isomorphism so that

lim−→(Mi ⊗N )∼=

lim−→Mi

⊗N .

First we show the direct limit of (Mi ×N , µi j × idN ) is M ×N . Indeed, let αi : Mi ×N → Q be a collection ofA-linear maps such that for all i ≤ j we have αi = α j (µi j × idN ). For M ×N to satisfy the universal propertycharacterizing lim−→(Mi ×N ) ([2.16]), we want to define a unique α : M ×N →Q such that αi = α (µi × idN ). This

36


forces us to attempt the definition α(µi (xi ), y) := αi (xi , y). Now α is defined on all of M ×N since by [2.15] eachelement x ∈ M is µi (xi ) for some i ∈ I and xi ∈ Mi . To show it is well defined, suppose x = µi (xi ) = µ j (x j ). Thenthere is some k ≥ i , j , and x =µk (µi k (xi )) =µk (µ j k (x j )) since µi =µk µi k . Now as αi = α j (µi j × idN )we have

α(µi (xi ), y) = αi (xi , y) = αk (µi k (xi ), y) = αk (µ j k (x j ), y) = α j (x j , y) = α(µ j (x j ), y),

so α is well defined. The existence of a unique such α shows that

lim−→Mi

×N ∼= lim−→

Mi ×N

.Letπi : Mi⊗N → P be the canonical map making P the direct limit, and for each i ∈ I let gi : Mi×N →Mi⊗N be

the canonical bilinear mapping. These gi form a homomorphism between the direct systems (Mi×N ) and (Mi⊗N ),so by [2.18] they induce a unique homomorphism

g : M ×N ∼= lim−→

Mi ×N

→ lim−→

Mi ⊗N

= P

such that g (µi × idN ) = πi gi . We have g (x, y) = gi (xi , y), and each gi is A-bilinear, so since any element ofM ×N has a representative in some Mi ×N , g is A-bilinear as well. Thus g induces an A-module homomorphismφ : M ⊗N → P such that

φ (µi ⊗ idN ) =πi (2.6)

for each i . Note on the other hand thatψ πi =µi ⊗ idN (2.7)

by the definition of ψ.Now by [2.15], each element p ∈ P can be written asπi (pi ) for some pi ∈Mi⊗N . Since this module is generated

by elements xi ⊗ y, for xi ∈ Mi and y ∈ N , by linearity of ψ and φ we may assume pi = xi ⊗ y. We then have, by[2.16], that

φ(ψ(p)) =φ(ψ(πi (pi )))Eq. 2.7= (φ (µi ⊗ idN ))(pi )

Eq. 2.6= πi (pi ) = p.

Similarly, let x⊗ y be a generator of M ⊗N . Then by [2.15] there are i ∈ I and xi ∈Mi such that x⊗ y =µi (xi )⊗ y,and

ψ(φ(x ⊗ y)) =ψ(φ[(µi ⊗ idN )(xi ⊗ y)])Eq. 2.6= ψ(πi (xi ⊗ y))

Eq. 2.7= (µi ⊗ idN )(xi ⊗ y) = x ⊗ y.

Thus ψ and φ are inverse isomorphisms.

21. Let (Ai )i∈I be a family of rings indexed by a directed set I , and for each pair i ≤ j in I let αi j : Ai →Aj be a ring homo-morphism, satisfying conditions (1) and (2) of Exercise 14 Regarding each Ai as a Z-module we can then form the directlimit A= lim−→Ai . Show that A inherits a ring structure from the Ai so that the mappings Ai →A are ring homomorphisms.The ring A is the direct limit of the system (Ai , αi j ).

If A= 0 prove that Ai = 0 for some i ∈ I .Let a, b ∈ A. By [2.15] there are i , j ∈ I and ai ∈ Ai , b j ∈ B j such that αi (ai ) = a and α j (b j ) = b . Now there is

k ∈ I such that k ≥ i , j , andαk (αi k (ai )) = αi (ai ) = a andαk (α j k (b j )) = α j (b j ) = b . Define ab = αk (αi k (ai )·α j k (b j )).We have made three choices in this definition, i , j , and k. Fixing i , j , suppose we picked k ′ instead of k. There issome l ∈ I with l ≥ k , k ′, and we have αk (αi k (ai ) · α j k (b j )) = (αl αk l )(αi k (ai ) · α j k (b j )) = αl (αi l (ai ) · α j l (b j )),and symmetrically for k ′, so the definition is independent of the choice of k. Now suppose instead we choose arepresentative b = α j ′(b j ′)with b j ′ ∈Aj ′ . Let k ≥ i , j , j ′. Consider c = α j k (b j )−α j ′k (b j ′). We chose these elementsso that αk (c) = 0, so by [2.15] there is l ≥ k such that αk l (c) = α j l (b j )−α j ′ l (b j ′) = 0, or α j l (b j ) = α j ′ l (b j ′). Takingk = l in the definition of ab , we see that the definition is independent of j . Symmetrically, it is independent of i .Thus we have a well defined multiplication on A.

Now by definition, if ai , bi ∈ Ai we have αi (ai )αi (bi ) = αi (αi i (ai )αi i (bi )) = αi (ai bi ), so the αi preserve multi-plication. To show they are ring homomorphisms, we just need to show αi (1) = 1 in A. But for any b ∈ A, we canwrite it as α j (b j ) for some b j ∈Aj , and pick k ≥ i , j , and then

αi (1)b = αk (αi k (1)α j k (b j )) = αk (α j k (b j )) = α j (b j ) = b

since αi k (1) = 1, each αi k being a ring homomorphism. Thus each αi is a ring homomorphism.To verify the ring axioms for A, we just need to note that for any three elements a, b , c ∈ A the elements

ab , ba, a(b c), (ab )c , ab +ac , a(b + c) are calculated via representatives in some ring Ak , then sent into A via αk ;since they hold in each Ak , they hold in A.

37


We can in fact prove A = 0 ⇐⇒ ∃i ∈ I (Ai = 0). Assume A = 0; then 1 = 0 in A. Now since αi : Ai → A isa ring homomorphism, αi (1) = 1. By [2.15], there is j ≥ i such that αi j (1) = 0. But the αi j are defined to be ringhomomorphisms, in particular sending 1 to 1. Thus 1= 0 in Aj , so Aj = 0.

On the other hand, if some Ai = 0, then for all j ≥ i we have 1= αi j (1) = αi j (0) = 0, so Aj = 0 for j ≥ i . Nowany element a ∈ A can be written as αk (ak ) for some k and ak ∈ Ak , by [2.15]. Find j ≥ i , k; then a = αk (ak ) =α j (αk j (ak )) = α j (0) = 0, so A= 0.

22. Let (Ai , αi j ) be a direct system of rings and let Ni be the nilradical of Ai . Show that lim−→Ni is the nilradical of lim−→Ai .If each Ai is an integral domain, then lim−→Ai is an integral domain.Let A= lim−→Ai and suppose a is in its nilradical. Then there is n > 0 such that an = 0. Let i ∈ I and ai ∈ Ai be

such that αi (ai ) = a. Then 0 = an = αi (ai )n = αi (a

ni ), so by [2.15] there is j ≥ i such that αi j (a

ni ) = αi j (ai )

n = 0.Write a′ = αi j (ai ) ∈ Aj . Then a′ ∈ N j and α j (a

′) = α j (αi j (ai )) = αi (ai ) = a, so a ∈ lim−→Ni . On the other hand, ifan

j = 0, then surely α j (a j )n = 0, so lim−→N j is contained in the nilradical of A.

Similarly, suppose A is not an integral domain. Then there exist nonzero a, b ∈ A such that ab = 0. Sincea, b 6= 0, by [2.15], no representative of a or b can be zero. Let αi (ai ) = a and α j (b j ) = b , and find k ≥ i , j . Thenab = µk (αi k (ai )α j k (b j )) = 0, so by [2.15] there is l ≥ k such that αl k (αi k (ai )α j k (b j )) = αi l (ai )α j l (b j ) = 0. Buta = αl (αi l (ai )) and b = αl (α j l (b j )) are nonzero, so αi l (ai ) and α j l (b j ) are nonzero, hence zero-divisors in Al .

23. Let (Bλ)λ∈Λ be a family of A-algebras. For each finite subset of Λ let BJ denote the tensor product (over A) of the Bλ forλ ∈ J . If J ′ is another finite subset of Λ and J ⊆ J ′, there is a canonical A-algebra homomorphism BJ → BJ ′ . Let B denotethe direct limit of the rings BJ as J runs through all finite subsets of Λ. The ring B has a natural A-algebra structure forwhich the homomorphisms BJ → B are A-algebra homomorphisms. The A-algebra B is the tensor product of the family(Bλ)λ∈Λ.

We should first note that the map given on p. 31 making the tensor product of A-algebras an A-algebra is amisprint. If we have ring homomorphisms f : A→ B and g : A→C , and define h ′ : A→ B⊗AC by h ′(a) = f (a)⊗g (a),then h ′(1) = 1⊗ 1 but

h ′(a) = f (a)⊗ g (a) = a2(1⊗ 1) 6= a(1⊗ 1) = ah ′(1)

in general. The proper definition is instead h : a 7→ f (a)⊗ 1= 1⊗ g (a) = a(1⊗ 1).If J = λ1, . . . , λm and J ′ = J ∪ λm+1, . . . , λn, the canonical homomorphism BJ → BJ ′ is given by bλ1

⊗ · · · ⊗bλm7→ bλ1

⊗· · ·⊗bλm⊗1⊗· · ·⊗1. It is obviously an A-algebra homomorphism. LetβJ : BJ → B denote the canonical

map associated to the direct limit. The A-algebra structure on B can be given as follows: let b ∈ B and a ∈ A. Thereare a finite J ⊆ Λ and an element bJ ∈ BJ such that b = βJ (bJ ); define ab = βJ (abJ ). Since the maps BJ → BJ ′

are A-algebra homomorphisms, this definition is independent of the choice of J . We saw in [2.21] that βJ is a ringhomomorphism, and by the definition of scalar multiplication in B it is also an A-algebra homomorphism.

Flatness and TorIn these Exercises it will be assumed that the reader is familiar with the definition and basic properties of the Tor functor.

24. If M is an A-module, the following are equivalent:i) M is flat;ii) TorA

n (M , N ) = 0 for all n > 0 and all A-modules N ;iii) TorA

1 (M , N ) = 0 for all A-modules N .i) =⇒ ii): Recall that TorA

n (M , −) are the derived functors of −⊗AM . This means that if we let

P : · · · → P2→ P1→ P0→N → 0

be a projective resolution of N , and tensor with M to get

P ⊗AM : · · · → P2⊗AM → P1⊗AM → P0⊗AM → 0

(lopping off the M ⊗N term so that we get H0 =M ⊗N ), then the homology groups Hn(P ⊗AM ) = ker(Pn ⊗M →Pn−1⊗M )/ im(Pn+1⊗M → Pn ⊗M ) are by definition the groups TorA

n (M , N ). Since M is flat, the sequence P ⊗AMis exact except at P0⊗AM , so the homology groups TorA

n (M , N ) = 0 for n > 0.

38


ii) =⇒ iii): 1> 0.iii) =⇒ i): Let 0 → N ′ → N → N ′′ → 0 be a short exact sequence of A-module homomorphisms. Since the

TorAn (M , −) are derived functors, they fit into a Tor exact sequence including TorA

1 (M , N ′′)→ M ⊗AN ′→ M ⊗AN .As by assumption TorA

1 (M , −) = 0, we get a short exact sequence 0→M ⊗AN ′→M ⊗AN ; as the injection N ′Nwas arbitrary, M is flat by (2.19).

25. Let 0→N ′→N →N ′′→ 0 be an exact sequence, with N ′′ flat. Then N ′ is flat ⇐⇒ N is flat.The book suggests we use the Tor exact sequence. It seems that this requires us to use the additional fact (not a

priori obvious) that TorAn (M , N ) ∼= TorA

n (N , M ) for all n ≥ 0 and A-modules M , N . Making this assumption, let Mbe an arbitrary A-module; we have an exact sequence

0 0

· · · // TorA2 (M ,N ′′) // TorA

1 (M ,N ′) // TorA1 (M ,N ) // TorA

1 (M ,N ′′) // M ⊗AN ′ // · · · .The criterion [2.24.iii] and the isomorphism TorA

1 (N′, M )∼=TorA

1 (N , M )mean N is flat just if N ′ is.We now prove that TorA

n (M , N )∼=TorAn (N , M ).18

First, a lemma: if F is a free A-module, then TorA1 (F , −) = 0 and TorA

1 (−, F ) = 0. Since F is flat by [2.4], by [2.24]we have TorA

1 (F , −) = 0. As for TorA1 (−, F ),19 consider the free resolution P2 = 0→ P1 = 0→ P0 = F → F → 0 of

F ; tensoring with any A-module M we get a sequence 0→ P1⊗M = 0→ F ⊗M → 0, whose homology TorA1 (M , F )

at P1⊗M = 0 is 0.Now let M and N be arbitrary A-modules. We can write

them as quotients of free A-modules F , G, so that we have exactsequences D : 0→ M ′ → F → M → 0 and E : 0→ N ′ → G →N → 0. By [2.4], F and G are flat, so the sequences 0→ M ′ ⊗G → F ⊗ G → M ⊗ G → 0 and 0 → F ⊗ N ′ → F ⊗ G →F ⊗N → 0 are exact. The Tor exact sequence for M ⊗ E is 0 =TorA

1 (M , G)→TorA1 (M , N )→M⊗N ′→M⊗G→M⊗N → 0.

Tensoring E with D , adding in the Tor exact sequence, and using(2.18), we have the commutative diagram on the right with exactrows and columns. The Snake Lemma, applied to the middle tworows, gives an exact sequence 0 → TorA

1 (M , N ) → M ′ ⊗N →F ⊗N . On the other hand, the Tor exact sequence for N ⊗ Dand commutativity of the tensor product (2.14.i) give an exactsequence

TorA1 (M , G)

0

0 //

TorA1 (M , N )

//

M ′⊗N ′ //

F ⊗N ′ //

M ⊗N ′ //

0

0 // M ′⊗G //

F ⊗G //

M ⊗G //

0

M ′⊗N //

F ⊗N //

M ⊗N //

0

0 0 0

0=TorA1 (N , F )→TorA

1 (N , M )M ′⊗N → F ⊗N .

Since TorA1 (M , N ) and TorA

1 (N , M ) both embed as the kernel of M ′⊗N → F⊗N , we see TorA1 (M , N )∼=TorA

1 (N , M ).

26. Let N be an A-module. Then N is flat ⇐⇒ Tor1(A/a, N ) = 0 for all finitely generated ideals a in A.The implication =⇒ follows from [2.24].For⇐=, assume Tor1(A/a, N ) = 0 for all finitely generated ideals a in A. Then given E : 0→ a→A→A/a→ 0,

the Tor sequence of E ⊗N shows that a⊗N →A⊗N is injective. Now let b be an arbitrary ideal of A; we want toshow b⊗N → A⊗N injective. Inclusions ai ,→ a j of finitely generated A-submodules (ideals) in b induce maps ofexact sequences

0 // ai ⊗N //

_

A⊗N // // A/ai ⊗N //

0

0 // a j ⊗N // A⊗N // // A/a j ⊗N // 0,

which piece together to give homomorphisms between the exact systems (ai⊗N ), (A⊗N ), and (A/ai⊗N ) for finitelygenerated ideals ai ⊆ b of A. By [2.17], the direct limit lim−→ai = b. Obviously lim−→A= A, and similarly, lim−→A/ai

∼=A/b.20 By [2.20]we have lim−→(ai⊗N )∼= b⊗N and lim−→(A⊗N ) =A⊗N ∼=N , and lim−→(A/ai⊗N )∼=A/b⊗N ∼=N/bN

18 http://uni.edu/ajur/v3n3/Banerjee%20pp%207-14.pdf19 http://math.uchicago.edu/~may/MISC/TorExt.pdf20 To see this we show that A/b has the universal property ([2.16]) of the direct system (A/ai , πi j ), writingπi j : A/ai A/a j andπi : A/ai

A/b. Set π0i : A A/ai . Evidently π j πi j = πi : A/ai A/a j A/b. Let αi : A/ai → P be such that α j πi j = αi . If we want to define

39

http://uni.edu/ajur/v3n3/Banerjee%20pp%207-14.pdf

http://math.uchicago.edu/~may/MISC/TorExt.pdf


by [2.2]. Then [2.19] states that direct limit is an exact functor, so the direct limit 0→ b⊗N →A⊗N →N/bN → 0is also exact, and thus b⊗N → A⊗N is injective. Writing E ′ : 0 → b → A→ A/b → 0 and looking at the Torsequence of E ′⊗N , the injectivity of this map shows Tor1(A/b, N ) = 0.

Now let M be a finitely generated module, generated by say x1, . . . , xn , and for i = 0, . . . , n let Mi =∑i

j=0 Ax j .Then each Mi/Mi−1, i = 1, . . . n is generated by one element xi , so the map fi : A→ Mi/Mi+1 given by a 7→ axi is asurjective A-module homomorphism, and by the second display of p. 19 (the first isomorphism theorem), if we writeai = ker( fi ) we have A/ai

∼= Mi/Mi−1. Consider the short exact sequences Ei : 0→ Mi−1 → Mi → A/ai → 0 givenby these isomorphisms. Suppose inductively that Tor1(Mi−1, N ) = 0; this is trivial for i = 1 and M0 = 0 = A/(1).The Tor sequence of Ei ⊗N gives, in part, 0=Tor1(Mi−1, N )→Tor1(Mi , N )→Tor1(A/ai , N ) = 0, so by exactnessTor1(Mi , N ) = 0. By induction, Tor1(M , N ) = 0 for all finitely generated modules.

Now let 0→ K → P be any injection of finitely generated A-modules. Complete this to a short exact sequenceE ′′ : 0 → K → P → K/P → 0, where K/P is finitely generated as well. The Tor sequence of E ′′ ⊗N gives 0 =Tor(K/P, N )→K ⊗N → P ⊗N , so K ⊗N → P ⊗N is injective. Since K P was an arbitrary injection of finitelygenerated A-modules, by criterion iv) of (2.19), N is flat.

27. A ring A is absolutely flat if every A-module is flat. Prove that the following are equivalent:i) A is absolutely flat;ii) every principal ideal is idempotent;iii) every finitely generated ideal is a direct summand of A.

i) =⇒ ii): Let a Ã A. The inclusion a ,→ A is of course injective; by assumption, the module A/a is flat, so theinduced map a⊗AA/a→ A⊗AA/a

∼−→ A/a is injective, where the isomorphism is by the absorption law (2.14.iv).But this composition is the zero map, for it takes a⊗ 1 7→ a⊗ 1 7→ a = 0, and thus the module a⊗AA/a = 0. Sincewe have a short exact sequence 0 → a → A→ A/a → 0, tensoring with a, assumed flat, gives by (2.19), an exactsequence 0= a⊗Aa→A⊗Aa→ 0= a⊗AA/a, showing the composition a⊗Aa→A⊗Aa→ a, using (2.14.iv), is anisomorphism. But this sends a⊗ a′ 7→ a⊗ a′ 7→ aa′, so since this map is surjective, every element of a is a finite sumof elements aa′ with a, a′ ∈ a and thus a= a2. In particular, for any x ∈A we have (x) = (x)2 idempotent.

ii) =⇒ iii): If every finitely generated ideal a is generated by some idempotent e , then by the proof of iii) =⇒ii) in [1.22], we have a decomposition A∼= (e)⊕ (1− e). Obviously this decomposition is only interesting if e 6= 0, 1.

It remains to show each finitely generated ideal is generated by a single idempotent. For a principal ideal (x),by assumption x ∈ (x2), so we may write x = ax2 for some a ∈ A. Multiplying both sides by a, we have ax =a2x2 = (ax)2, so e = ax is idempotent. Since e = ax ∈ (x) and x = ax2 = (ax)x = e x ∈ (e), we have (x) =(e). Now any finitely generated ideal a = (x1, . . . , xn) = (e1, . . . , en) is generated by idempotents, where ei is anidempotent generating (xi ). As in [1.11.iii], we show every ideal finitely generated by idempotents is generated bya single element. This is trivial for n = 1, so inductively suppose it holds for all ideals with n generators, and leta= (x1, . . . , xn , y) be an ideal generated by n+ 1 elements. Let e be an idempotent generating (x1, . . . , xn) and f anidempotent generating (y), so that a = (e , f ). If we let z = e + f − e f , then we have e z = e2 + e f − e2 f = e andf z = f e + f 2 − f e f = f , so a = (e , f ) = (z) is principal, and there is thus an associated idempotent g such that(g ) = (z) = a.

iii) =⇒ i): Let N be an arbitrary A-module. To show it is flat, by [2.26] it suffices to show that TorA1 (A/a, N ) = 0

for all finitely generated ideals a ∈ A. By assumption, each of these is a direct summand, so A ∼= a⊕A/a, and by(2.14.iv,iii) we have isomorphisms N ∼= A⊗N ∼= (a⊗N )⊕ (A/a⊗N ), so the inclusion a ,→ A induces an injectiona⊗N A⊗N . Now the Tor exact sequence for 0→ a→ A→ A/a→ 0 includes the fragment 0= TorA

1 (A, N )→TorA

1 (A/a, N )→ a⊗N →A⊗N , so TorA1 (A/a, N ) is isomorphic to the kernel of a⊗N A⊗N , which is zero.

28. A Boolean ring is absolutely flat.In a Boolean ring each element is idempotent, so each principal ideal is idempotent, so by [2.27] the ring is

absolutely flat.

The ring of Chapter 1, Exercise 7 is absolutely flat.Recall that this is a ring A in which for every element a there is a number n = n(a)> 1 such that an = a. Then

a = a2an−2 ∈ (a2) so every principal ideal is idempotent, and by [2.27] A is absolutely flat.

α : A/b→ P such that απi = αi , taking i = 0, we are forced to try α(π0(a)) = α(a+b) = α0(a) for a ∈A. Indeed, for any i and any a+ai ∈A/aiwe have

αi (a+ ai ) = αi (π0i (a)) = α0(a) = α(π0(a)) = α(πi (a+ ai ))so this homomorphism meets the requirement and A/b has the universal property.

40


CITE [1.12]?

Every homomorphic image of an absolutely flat ring is absolutely flat.Let A be absolutely flat and let (x) be a principal ideal in A/a. Then for the lift x ∈ A we have, by [2.27], that

(x)2 = (x), so there is a ∈ A such that ax2 = x. Downstairs in A/a we have a x2 = x, so (x)2 = (x) and A/a isabsolutely flat by [2.27] again.

If a local ring is absolutely flat, then it is a field.Let m be the maximal ideal of a local, absolutely flat ring A. We want to show m = 0. Suppose x ∈m. Then by

[2.27]we have an idempotent e ∈ (x)with (e) = (x), and e = 0 ⇐⇒ x = 0. Then f = 1− e is an idempotent as well,but also f is a unit by (1.9) since e is in the Jacobson radical R=m. Then we have 1= f −1 f = f −1 f 2 = ( f −1 f ) f = f ,so e = 0. Thus m= 0, so A∼=A/(0) =A/m is a field.

If A is absolutely flat, every non-unit in A is a zero-divisor.Let x ∈A be a non-unit; then (x) 6= (1) is a finitely generated ideal, and so by [2.27] there is another ideal b 6= (0)

such that A∼= (x)⊕ b. Then bx ∈ (x)∩ b= (0), so x is a zero-divisor.

41

Chapter 3: Rings and Modules of Fractions

Exercise. Verify that these definitions [repeated below] are independent of the choices of representatives (a, s) and (b , t ),and that S−1A satisfies the axioms of a commutative ring with identity.

We recall that S is a multiplicative submonoid of A, meaning a subset closed under multiplication and containing1. An element of S−1A is defined to be an equivalence class a/s of pairs (a, s) ∈A× S under the relation ≡ given by

(a, s)≡ (b , t ) ⇐⇒ ∃u ∈ S [(at − b s)u = 0].

The book shows that ≡ indeed is an equivalence relation, and then defines

as+

bt

:=at + b s

st,

as· b

t:=

asb s

.

It falls to us to verify these operations are well defined. To show as +

bt is independent of the representatives of a/s

and b/t chosen, suppose we calculated with two pairs of representatives (a, s) ≡ (a′, s ′) and (b , t ) ≡ (b ′, t ′). Thenby definition there are u, v ∈ S such that (as ′− a′s)u = 0= (bt ′− b ′t )v in A. We need to verify that (at + bs , st )≡(a′t ′+ b ′s ′, s ′ t ′), meaning that there exists w ∈ S such that ([at + b s]s ′t ′− [a′t ′+ b ′s ′]s t )w = 0 in A. But w = uvworks, for

w([at+b s]s ′t ′−[a′t ′+b ′s ′]s t ) = uv(as ′tt ′+b ss ′t ′−a′stt ′−b ′ss ′t ) = u(as ′−a′s)tt ′v+v(bt ′−b ′t )ss ′u = 0+0= 0.

Similarly, · is well defined: (ab , st )≡ (a′b ′, s ′t ′), for setting w = uv we have

w(ab s ′t ′− a′b ′st ) = uv(ab s ′t ′− a′bst ′+ a′b st ′− a′b ′st ) = u(as ′− a′s)b t ′v + v(b t ′− b ′t )a′su = 0+ 0= 0.

Note as a preliminary that for any a/s ∈A and t ∈ S we have at/st = a/s , for (at )s−a(st ) = 0 by commutativityand associativity of · in A.

Now we verify that (S−1A, +, 0/1) is an abelian group. + is associative, since given a/s , b/t , c/u ∈ S−1A wehave

as+

bt

+cu=

at + b sst

+cu=

atu + b su + cststu

=as+

bu + cttu

=as+

bt+

cu

,

using distributivity and commutativity of · in A. We see + is commutative, for

as+

bt=

at + b sst

=b s + at

ts=

bt+

as

,

using commutativity of + and · in A. For any element v ∈ S, we have 0/v = 0v/1v = 0/1 a neutral element for +,since

01+

as=

0s + a11s

=as

,

using the properties of 0 and 1 in A. The additive inverse of a/s is (−a)/s , because

as+−as=

as +(−a)ss2

=0s2=

01

.

Now we want to prove that (S−1A, +, ·, 0/1, 1/1) is a commutative ring. The new · is associative since

as

bt

cu=

abst

cu=

abcstu=

as

bctu=

as

bt

cu

,

42

Chapter 3: Rings and Modules of Fractions Ex. 3.1

implicitly using associativity of · in A. The new · is commutative because

as

bt=

abst=

bats=

bt

as

,

using commutativity of · in A. The element 1/1 is neutral for · because

11

as=

1a1s=

as

,

1 being neutral for · in A. Finally, · distributes over + because

as

bt+

cu

=as

bu + cttu

=abu + act

stu=

absu + acststsu

=absustsu

+acststsu

=abst+

acsu=

as

bt+

as

cu

.

Let M be an A-module. If we redefine a, b , c to be elements of an M , then (but for our tendency to write as ratherthan sa, which strictly speaking doesn’t matter for modules over commutative rings) the proofs of well-definedness,associativity, commutativity, 0/1, and −a/s = −(a/s) for (S−1A, +, 0/1) go through to define an abelian groupstructure on (S−1M , +, 0/1). Now we show S−1M carries a natural S−1A-module structure. If we let b = m ∈M , thewell-definedness proof for · in S−1A shows that a

smt =

amst gives a well-defined scalar product S−1A× S−1M → S−1M .

We have the four axioms of p. 17 to verify. If we let a ∈M , the proof 11

as =

as shows 1/1 acts as the identity on S−1M ,

the fourth axiom. Letting b , c ∈ M , the associativity of ·, and the distributivity of · over + for S−1A provide thethird and first axioms. Letting a, b ∈A, s , t , u ∈ S and m ∈M , the second (and last) axiom is

as+

bt

mu=

at + bsst

mu=

atm+ bsmstu

=atmstu+

bsmstu=

amsu+

bmtu

.

Proposition 3.11. v) The operation S−1 commutes with formation of finite sums, products, intersections and radicals.(3.4.i,ii) show S−1 distributes over finite sums and intersections. (1.18) shows S−1(ab) = (S−1a)(S−1b), and

S−1 r (a) ⊆ r (S−1a). It remains to show r (S−1a) ⊆ S−1 r (a), so suppose x/s ∈ S−1A is in r (S−1a). Then for somen > 0 its nth power is some a/t ∈ S−1a, meaning xn/s n = (x/s)n = a/t in S−1A. By the definition of equalityin S−1A, there is some u ∈ S so that utxn = us na ∈ a. Multiplying both sides by (ut )n−1 we see (utx)n ∈ a, andutx ∈ r (a). Thus utx/uts = x/s ∈ S−1 r (a) as claimed.

EXERCISES1. Let S be a multiplicatively closed subset of a ring A, and let M be a finitely generated A-module. Prove that S−1M = 0 if

and only if there exists s ∈ S such that s M = 0.Assume s M = 0 for some s ∈ S, and let m/t ∈ S−1M . Then s(1m− t0) = 0 in M , so m/t = 0/1 in S−1M .Conversely, let M be an A-module finitely generated by m1, . . . , mn , and suppose that S−1M = 0. Then in par-

ticular we have for each i that m/1 = 0/1 in S−1M , so there is si ∈ S such that 0 = si (1mi − 1 · 0) = si mi in M .Let s = s1 · · · sn , which is in S since S is multiplicatively closed. Then for any element m =

∑

ai mi ∈ M we have0= s m = s(1m− 1 · 0), and m/1= 0/1 in S−1M .

2. Let a be an ideal of a ring A, and let S = 1+ a. Show that S−1a is contained in the Jacobson radical of S−1A.Since 0 ∈ a we have 1 ∈ S = 1+ a, and if a, b ∈ a, then (1+ a)(1+ b ) = 1+ a + b + ab ∈ S = 1+ a, so

S is multiplicatively closed. Now to show S−1a ⊆ R(S−1A), it is enough, by (1.9), to show the set 1 − S−1a =1− (S−1A)(S−1a) is made up of units. But if a

1+b ∈ S−1a, then 11 −

a1+b =

1+b−a1+b ∈ S−1S ⊆ (S−1A)×.

Use this result and Nakayama’s lemma to give a proof of (2.5) which does not depend on determinants.(2.5) states that for all finitely generated A-modules M and aÃ A such that aM = M there is x ∈ 1+ a such that

xM = 0.So suppose M is finitely generated by some m1, . . . , mn with aM =M . Then localizing by S = 1+ a, by (3.11.v)

we have (S−1a)(S−1M ) = S−1M . The last paragraph shows that S−1a⊆R(S−1A), the Jacobson radical, and S−1M isfinitely generated over S−1A by m1/1, . . . , mn/1, so the conditions of Nakayama’s Lemma are met, and we concludeS−1M = 0. But then by [3.1] there is x ∈ S = 1+ a such that xM = 0.

43

Ex. 3.3 Chapter 3: Rings and Modules of Fractions

3. Let A be a ring, let S and T be two multiplicatively closed subsets of A, and let U be the image of T in S−1A. Show thatthe rings (ST )−1A and U−1(S−1A) are isomorphic.

As a preliminary, we prove the canonical map φS : A→ S−1A is a epimorphism, meaning not that is surjective,but that it is right-cancellable. Suppose we have a ring homomorphisms µ : S−1A→ B , and let λ = µ φS . Sincefor all s ∈ S we have φS (s) ∈ S−1A a unit, we then have λ(s) = µ(φS (s)) a unit of B . By (3.1) there is a unique ringhomomorphism ν : S−1A→ B such that ν φS = λ. Thus

µ φS = ν φS =⇒ µ= ν . (3.1)

Now U is a multiplicative submonoid of S−1A since it is the image of a multiplicative submonoid under a ringhomomorphism, which preserves multiplication and unity. ST is also multiplicative submonoid of A, since 1 ∈ S, Timplies 1 = 1 · 1 ∈ ST , and if s t , s ′ t ′ ∈ ST , where s , s ′ ∈ S and t , t ′ ∈ T , then (s t )(s ′ t ′) = (s s ′)(t t ′) ∈ ST since Sand T are multiplicatively closed. Note S = S · 1⊆ ST and T = 1T ⊆ ST .

Consider the canonical map φST : A→ (ST )−1A. Each element of S is taken to a unit, since S ⊆ ST , so by (3.1)there is a unique homomorphism ρST

S : S−1A→ (ST )−1A such that

ρSTS φS =φST . (3.2)

AφS //

φST ((

S−1AφU //

ρSTS

##

U−1(S−1A)

ψ

(ST )−1A

ψ′

OOBy the proof of that proposition we have ρST

S (a/s) = φST (a)φST (s)−1 = a/s ∈

(ST )−1A. Now each element t/1 ∈ U ⊆ S−1T ⊆ S−1A is taken by ρSTS to t/1 ∈

(ST )−1A, which is a unit since T ⊆ ST . Then (3.1) again induces a unique homo-morphism ψ : U−1(S−1A) → (ST )−1A such that, if φU : S−1A → U−1(S−1A) is thecanonical map, then

ψ φU = ρSTS . (3.3)

Composing with φS : A→ S−1A on the right we get

ψ φU φSEq. 3.3= ρST

S φSEq. 3.2= φST . (3.4)

If ψ is a bijection, we are done. However it’ s more natural to use universal properties to construct an inverse.Now the composition φU φS : A→ S−1A→ U−1(S−1A) taking a 7→ a/1 7→ a/1

1/1 takes each element of S to a unit

(inverse 1/s1/1 ) and each element of T to a unit (inverse 1/1

t/1 ), and so takes each element of ST to a unit. By (3.1), thereis a unique homomorphism ψ′ : (ST )−1A→U−1(S−1A) such that

ψ′ φST =φU φS . (3.5)

Composing with ψ on the left gives ψ ψ′ φSTEq. 3.5= ψ φU φS

Eq. 3.4= φST . Since φST is an epimorphism (Eq.

3.1), ψ ψ′ = id(ST )−1A. On the other hand, since ψ φUEq. 3.3= ρST

S , composing with ψ′ on the left and φS on the

right gives ψ′ ψ φU φSEq. 3.4= ψ′ φST

Eq. 3.5= φU φS . Since φU and φS are epimorphisms (Eq. 3.1), we have

ψ′ ψ= idU−1(S−1A).1

1 A proof avoiding universal properties is as follows. Define a ring isomorphism ψ : (ST )−1A→ U−1(S−1A). Set ψ(a/s t ) = a/st/1 . To see it is

well-defined, suppose a/s t = a′/s ′ t ′ in (ST )−1A; we claim that a/st/1 =

a′/s ′

t ′/1 . This will follow if there is u = t0/1 ∈ U such that at ′ t0s = u a

st ′1 =

u a′

s ′t1 =

a′ t t0s ′ in S−1A. This will in turn be the case if there is s0 ∈ S such that s0at ′ t0 s ′ = s0a′ t t0 s in A. But since we assumed a/s t = a′/s ′ t ′, by

definition there is s ′′ t ′′ ∈ ST such that s ′′ t ′′as ′ t ′ = s ′′ t ′′a′s t , and we may take s0 = s ′′ and t0 = t ′′.Now we show it is a bijection. Any element of U−1(S−1A) can be written as a/s

t/1 for some a ∈ A, s ∈ S, t ∈ T , and then is mapped onto by

a/s t , so φ is surjective. Now suppose a/st/1 =φ(a/s t ) = 0= 0

1 =0/11/1 . By the definition of equality in U−1(S−1A), there is u = t ′/1 ∈U such that

at ′s =

t ′1

as

11 =

t ′1

01

t1 =

01 in S−1A. But then by the definition of equality in S−1A there is s ′ ∈ S such that s ′ t ′a = s ′0s = 0. Then a/1 = 0/1 in

(ST )−1A, so as t is zero and φ is injective.

It remains to show ψ is a ring homomorphism. Now ψ(1/(1 · 1)) = 1/11/1 is the unity of U−1(S−1A), and if we let a/s t and a′/s ′ t ′ ∈ (ST )−1A,

44


4. Let f : A→ B be a homomorphism of rings and let S be a multiplicatively closed subset of A. Let T = f (S). Show thatS−1B and T −1B are isomorphic as S−1A-modules.

As a homomorphic image of a multiplicative submonoid, T is a multiplicative submonoid of B . Now g = idB × fis a surjection B×S B×T , and we will show it induces a bijection of equivalence classes S−1B↔ T −1B . The actionof S ⊆A on B , by definition is s · b = f (s)b , so the equivalence relation ≡S on B × S is defined by (b , s)≡S (b

′, s ′) :⇐⇒ ∃s ′′ ∈ S ( f (s ′′s)′b = f (s ′′s)b ′) this is the same equation that holds just if (b , f (s))≡T (b

′, f (s ′)) in B×T . Thustwo elements of B× S define the same element of S−1B just if their g -images define the same element of T −1B , so ginduces a bijection φ : S−1B→ T −1B . Since f is a homomorphism, it follows easily that φ is also a homomorphism

of rings. Finally, if a/s ∈ S−1A and b/s ′ ∈ S−1B , we have φ

as

bs ′

=φ

f (a)bss ′

= f (a)bf (s) f (s ′) =

as

bf (s ′) =

asφ

bs ′

, so φ is a

S−1A-module isomorphism.

5. Let A be a ring. Suppose that, for each prime ideal p, the local ring Ap has no nilpotent element 6= 0. Show that A has nonilpotent element 6= 0. If each Ap is an integral domain, is A necessarily an integral domain?

By (3.12), we have N(Ap) = N(A)p for each prime p, where N(A) is the nilradical of A. By (3.8), this meansN(A) = 0.

It is possible for a ring with zero-divisors to have all localizations at primes integral domains, for suppose A=∏n

j=1 k j is a product of n ≥ 2 fields; it is not an integral domain, but we shall show its localizations at primes are.By [1.22], the only prime ideals of A are p j = 0k j ×

∏

i 6= j ki ; their complements are S j = k×j ×∏

i 6= j ki Now eachinserted k j is naturally an A-module, and A is a direct sum of these modules. By (3.4.i), localization distributes overfinite direct sums of A-modules. Now S−1

j k j∼= k j , while for i 6= j we have 0 ∈ S j · ki , so by [3.1], S−1

j ki = 0. Thusall localizations Ap j

∼= k j at primes are fields, and a fortiori integral domains.

6. Let A be a ring 6= 0 and let Σ be the set of all multiplicatively closed subsets S of A such that 0 /∈ S. Show that Σ hasmaximal elements and that S ∈Σ is maximal if and only if A\S is a minimal prime ideal of A.

Certainly 1 ∈ Σ, so Σ is non-empty. To find maximal elements of Σ, we apply Zorn’ s Lemma. Let (Sa)a∈Ibe a totally ordered chain in Σ; we claim its union S is an upper bound. Surely 0 /∈ S since 0 is in no Sa, and iftwo elements s , t ∈ S are given, they belong to some Sα and Sβ, respectively. If γ =maxα, β, then we have boths , t ∈ Sγ , so s t ∈ Sγ ⊆ S, and thus S ∈Σ is an upper bound for the chain.

Assume p is a prime ideal, and let S =A\p. Then by the definition of being prime, a, b ∈ S =A\p implies ab ∈ S,so S is multiplicatively closed. Conversely, if the complement of a multiplicative submonoid is an ideal, it is prime.Since 0 ∈ p, we don’t have 0 in S, so S ∈Σ.

Let S ∈Σ be maximal, and p=A\S. Note that the smallest multiplicative submonoid containing a ∈A and S issan : s ∈ S, n > 0. If a ∈ p, this monoid is strictly larger than S, and so by maximality of S ∈ Σ, contains zero.Thus a ∈ p just if there are n > 0 and s ∈ S with san = 0. Suppose a, b ∈ p, and let m, n > 0 and s , t ∈ S such thatsam = t b n = 0. If p = m + n − 1, then am or b n divides each term of (a − b )p so s t (a − b )p = 0, and a − b ∈ p.Thus p is an additive subgroup of A. If x ∈A is any other element, then s(ax)m = (sam)x m = 0x m = 0, so ax ∈ p aswell. Thus p is an ideal. If q( p was a smaller prime ideal, then A\q would be an element of Σ strictly containing S,which we assumed is impossible, so p is minimal.

If, on the other hand p is a minimal prime ideal, then S = A\p is an element of Σ. If T ⊇ S is maximal, thenA\T ⊂ p is a minimal prime ideal, hence equal to p=A\S, and so S = T is maximal.

7. A multiplicatively closed subset S of a ring A is said to be saturated if

xy ∈ S ⇐⇒ x ∈ S and y ∈ S.

we have the equations

φ

ast+

a′

s ′t ′

=φ

as ′t ′+ a′ststs ′t ′

=at ′s ′+a′ts

ss ′

tt ′1

=at ′s +

a′ts ′

tt ′1

=a s t ′

1 +a′

s ′t1

t1

t ′1

=a st1

+a′

s ′

t ′1

=φ

ast

+φ

a′

s ′ t ′

,

φ

ast· a′

s ′t ′

=φ

aa′

sts ′t ′

=aass ′

tt ′1

=as ·

a′

s ′

t1 ·

t ′1

=ast1

·a′

s ′

t ′1

=φ

ast

φ

a′

s ′t ′

.

45


Prove thati) S is saturated ⇐⇒ A\S is a union of prime ideals.

Suppose a subset S ⊆A is such that A\S =⋃

pα is a union of prime ideals. Then 1 /∈ pα for all α, so 1 ∈ S. Supposex, y ∈ S =A\

⋃

pα. Then for all pα we have x, y /∈ pα, so xy /∈ pα, and thus xy ∈⋂

(A\pα) =A\⋃

pα = S. Thus S isa multiplicative submonoid. On the other hand, if we have x /∈ S, then there is some pα 3 x, and that being an idealwe have xy ∈ pα ⊆A\S, and symmetrically for y. Thus S is saturated.

Now suppose S ⊆A is saturated. To show the complement is a union of prime ideals, it suffices to manufacture,for any element a ∈ A\S, a prime ideal p 3 a disjoint from S. Note that if a ∈ A\S, by saturation for all b ∈ A wehave ab /∈ S, so that (a) is an ideal disjoint from S. The set Υ of ideals of A containing a and disjoint from S is thennon-empty, and it is closed under increasing unions, so by Zorn’s Lemma it contains a maximal element p. We willbe done if we can show p is prime, so suppose x, y /∈ p. Then (x) + p and (y) + p are not in Υ , and so intersect S. Ifs , t ∈ S are such that s ∈ (x)+p and t ∈ (y)+p, then s t ∈

(x)+p

(y)+p

⊆ (xy)+p, so xy /∈ p. Thus p is prime.

ii) If S is any multiplicatively closed subset of A, there is a unique smallest saturated multiplicatively closed subset Scontaining S, and that S is the complement in A of the union of the prime ideals which do not meet S. (S is called thesaturation of S.)

Let S be the complement of the union of primes p not meeting S: S :=A\⋃

p ∈ Spec(A) : S ∩p=∅. Then S issaturated, by i), and contains S, since A\S ⊆ A\S. Moreover, any saturated set containing S is the complement of aunion of primes not meeting S, and since S is the complement of the largest such union, it is the smallest saturatedset containing S.

If S = 1+ a, find S.A prime p meets S just if we have a ∈ a and x ∈ p such that x = 1+ a, or 1 = a − x, so that (1) = a+ p. Thus

the union in A\S is over all prime ideals not coprime to a. In particular, for every such p, there is a maximal idealm ⊇ a+ p. Since every maximal ideal is prime and every prime ideal is contained in a maximal ideal, it suffices totake the union of maximal ideals containing a. Thus S =A\

⋃

m ∈Max(A) : a⊆m.

8. Let S, T be multiplicatively closed subsets of A, such that S ⊆ T . Let φ : S−1A→ T −1A be the homomorphism whichmaps each a/s ∈ S−1A to a/s considered as an element of T −1A. Show that the following statements are equivalent:i) φ is bijective.ii) For each t ∈ T , t/1 is a unit in S−1A.iii) For each t ∈ T there exists x ∈A such that x t ∈ S.iv) T is contained in the saturation of S (Exercise 7).v) Every prime ideal which meets T also meets S.

Note that S ⊆ T =⇒ ST = T since T is multiplicatively closed. Now use the unique homomorphismρT

S : S−1A→ T −1A, defined in the proof of [3.3], such that ρTS φS = φT . (3.1) shows ρT

S (a/s) = φT (a)φT (s)−1 =

a/s ∈ T −1A.i) =⇒ ii): If ρT

S is bijective, it is an isomorphism, so since ρTS (t/1) = t/1 ∈ T −1A is a unit (inverse 1/t ),

t/1 ∈ S−1A is also a unit.ii) =⇒ iii): If t/1 is a unit in S−1A, then there is x/s ∈ S−1A such that t x/1s = 1/1, which by definition means

there is s ′ ∈ S such that s ′tx1= s ′s1 in A. Then (s ′x)t ∈ S.iii) =⇒ i): Suppose ρT

S (a/s) = 0/1 in T −1A. Then there is t ∈ T such that t a = 0. If x ∈ A is such that x t ∈ S,then (xt)a = 0 shows that a/s = 0/1 in S−1A. Now let a/t ∈ T −1A be arbitrary, and let x ∈ A be such that x t ∈ S.Then a/t = xa/xt = ρT

S (xa/xt ) is the image of an element of S−1A.

iii) =⇒ iv): S is saturated, so if for each t ∈ T there is x ∈ A such that xt ∈ S ⊆ S, then by definition we havext ∈ S, so in particular T ⊆ S.

iv) =⇒ iii): Write S ′ = a ∈ A : ∃x ∈ A (ax ∈ S), so that by definition We claim S ′ = S is the saturation ofS. Surely S ⊆ S ′ ⊆ S, since if s ∈ S then s · 1 ∈ S, and since if ax ∈ S then ax ∈ S. To show the other inclusionit suffices to show S ′ is also saturated. Clearly, if ab ∈ S ′, then there exist xy ∈ A such that ax, by ∈ S, and thenab · xy = ax · by ∈ S, so ab ∈ S ′. Supposing on the other hand that a /∈ S ′, then there is no x ∈ A such that ax ∈ S,and certainly for all b , y ∈A we have aby /∈ S, so ab /∈ S ′; and symmetrically if b /∈ S ′. Thus S ′ = S is saturated.

By definition t ∈ S ′ ⇐⇒ ∃x ∈A (x t ∈ S).

46


iv) ⇐⇒ v):

T ⊆ S ⇐⇒ T ⊆ S = S ⇐⇒⋃

primes not meeting S=A\S ⊆A\T =⋃

primes not meeting T

⇐⇒ primes not meeting S ⊆ primes not meeting T ⇐⇒ primes meeting T ⊆ primes meeting S.

9. The set S0 of all non-zero-divisors in A is a saturated multiplicatively closed subset of A. Hence the set D of zero-divisorsin A is a union of prime ideals (see Chapter 1, Exercise 14). Show that every minimal prime ideal of A is contained in D.

If 1= 0, then S0 should probably be considered empty, so henceforward let’s assume not. Then 1 ∈ S0. If x, y ∈ S0,then xy 6= 0, and for all a ∈ A we have a(xy) = (ax)y 6= 0, so xy ∈ S0. Thus S0 is a multiplicative submonoid. Nowsuppose x ∈D , say with ax = 0. For any y ∈A we then have axy = 0, so xy ∈D ; thus S0 is saturated.

Recall from [3.6] that Σ is the collection of multiplicative submonoids S of A not containing 0. We claim that S0is contained in every maximal element S ∈Σ. Indeed, if we did not have S0 ⊆ S, then the product S0S would strictlycontain S, and thus contain s0 s = 0, for some s0 ∈ S0 and s ∈ S, contradicting the defining assumption S0 ∩D =∅.Now by [3.6] the maximal elements of Σ are of the form A\p for p a minimal prime of A, so we have A\D ⊆ A\p,or p⊆D , for all minimal primes p.

The ring S−10 A is called the total ring of fractions of A. Prove that

i) S0 is the largest multiplicatively closed subset of A for which the homomorphism A→ S−10 A is injective.

a 7→ a/1= 0/1 in S−1A implies there is some s ∈ S such that sa · 1= 0 · 1= 0 in A. This cannot happen if S ⊆ S0,but can happen for any S strictly larger than S, since such will contain a zero-divisor s .

ii) Every element in S−10 A is either a zero-divisor or a unit.

Let a/s ∈ S−10 A. If a/s is a zero-divisor, there is b/t ∈ S−1

0 A such that ab/st = 0/1, so there exists u ∈ S0 such thatuab = 0st = 0 in A, and ab = 0, then, since u is not a zero-divisor; thus a is a zero-divisor in A. Thus if a/s ∈ S−1

0 Ais not a zero-divisor, then a ∈ S0, so s/a ∈ S−1

0 A is an inverse to a/s , which is then a unit.

iii) Every ring in which every non-unit is a zero-divisor is equal to its total ring of fractions (i.e., A→ S−10 A is bijective).

If S = 1 we obviously have A∼= S−1A, and the inclusion 1 ,→ S0 induces the homomorphism φ : A→ S−10 A

as in [3.8]. This map is bijective just if, by condition ii), for each s ∈ S0, s/1 is a unit in S−10 A. But each s ∈ S0 has an

inverse s−1 in A by assumption, and then s−1/1 is an inverse of s/1 in S−10 A.

10. Let A be a ring.i) If A is absolutely flat (Chapter 2, Exercise 27) and S is any multiplicatively closed subset of A, then S−1A is absolutelyflat.

Let M be an S−1A-module, and write M |A for M viewed as an A-module by restriction of scalars along the canon-ical map A→ S−1A. We can then take S−1(M |A), allowing division by elements of S again, and we want to show thecomposition of natural mapsψ : M →M |A→ S−1(M |A) taking m 7→ m 7→ m/1 gives an S−1A-module isomorphism.For surjectivity, let m/s be any element of S−1(M |A). In M , the scalar product m′ = 1

s m is defined, and we havesm′ = m in the module M . Since s ∈A, we also have sm′ = m in M |A. But then, by definition m′/1= m/s in S−1M ,so ψ is surjective. For injectivity, suppose ψ(m) = m/1= 0/1 in S−1(M |A). Then there is s ∈ S such that sm = 0 inM |A. But then sm = 0 in M , so 0 = 1

s sm = m. Thus ψ is injective. That ψ preserves the S−1A-module structure isseen as follows. All homomorphisms are A-linear. If we take m ∈ M and apply 1/s to get m′ = 1

s m, then we haves m′ = m in M |A, since s ∈A, and thus s(m′/1) = sψ(m′) = m in S−1(M |A). But we also have s m

s = m in S−1(M |A),so s

m′

1 −ms

= 0 ∈ S−1(M |A), and multiplying on the left by 1/s we see m′/1= m/s . Thus ψ

1s m

= 1sψ(m), so ψ

is S−1A-linear.To show S−1A is absolutely flat, now, let M be an S−1A module and φ : N ′ N an injective S−1A-module

homomorphism. We want idM ⊗φ : M⊗S−1AN ′M⊗S−1AN to be injective. Note that M |A⊗AN ′|A→M |A⊗AN |A isinjective since all A-modules are flat. Since localization is exact, we also have S−1(M |A⊗AN ′|A)→ S−1(M |A⊗AN ′|A)injective. But by (3.7), this is equivalent to an injective S−1A-module homomorphism S−1(M |A)⊗S−1AS−1(N ′|A)S−1(M |A)⊗S−1AS−1(N |A), and we have shown that this is the same as idM ⊗φ : M ⊗S−1AN ′M ⊗S−1AN .

47


ii) A is absolutely flat ⇐⇒ Am is a field for each maximal ideal m.If A is absolutely flat, then by i) each Am is absolutely flat. But then by [2.28], Am, being local, is a field.Now assume each localization Am is a field, and let M be an A-module. The localization Mm is an Am-module,

and since Am is a field, it is a free Am-module. But free modules are flat, by [2.4] (sums of flat modules are flat, andvice versa), and the absorption law (2.14.iv). Thus Mm is a flat Am-module for each m. By (3.10), M is a flat A-module.

11. Let A be a ring. Prove that the following are equivalent:i) A/N is absolutely flat (N being the nilradical of A).ii) Every prime ideal of A is maximal.iii) Spec(A) is a T1-space (i.e., every subset consisting of a single point is closed).iv) Spec(A) is Hausdorff.

i) =⇒ iv): Let A/N be absolutely flat and X = Spec(A/N). Let x 6= y ∈ X be two distinct points. We findthem disjoint basic open neighborhoods (defined in [1.17]) Xe , Xf . Since px , py are distinct maximal ideals, we havepx + py = (1), so there are elements a ∈ px and b ∈ py with (a) + (b ) = (1). By [2.27.ii], there are idempotents egenerating (a) and g generating (b ), so that (e , g ) = (1). Let f = g (1− e). Then e f = 0, while g = e g + f ∈ (e , f ),so (e , f ) = (1). Since e ∈ px and px 6= (1), we have x ∈Xf , and similarly y ∈Xe . But by [1.17.i,ii] we have Xe ∩Xf =Xe f = X0 = ∅. Also Xe ∪Xf = V (e)∩V ( f ) = V

(e) + ( f )

= V (1) = ∅. Now Spec(A) is homeomorphic to X by[1.21.iv], and so also Hausdorff.

iv) =⇒ iii): Fix x ∈X. For each y 6= x we have a Uy containing y but not x. Then x=X \⋃

y 6=x Uy is closed.iii) ⇐⇒ ii): By [1.18.i], x is closed just if px is maximal. Thus all singletons are closed just if all primes are

maximal.ii) ⇐⇒ i): All primes of A are maximal

(1.7)⇐⇒∀m ∈Max(A), no prime ideal of A is strictly between m and N(1.1)⇐⇒∀m ∈Max(A), the only prime of A/N contained in m/N is (0)

(3.11.iv)⇐⇒∀m ∈Max(A), the only prime ideal of (A/N)m is (0)⇐⇒∀m ∈Max(A), (A/N)m is a field[3.10]⇐⇒ A/N is absolutely flat.

If these conditions are satisfied, show that Spec(A) is compact and totally disconnected (i.e., the only connected subsetsof Spec(A) are those consisting of a single point).

We already showed Spec(A) was compact in [1.17]. In the proof that i) =⇒ iv) above, we found, for any twodistinct points x, y, disjoint open neighborhoods Xf , Xe whose union is the entire space. Any subset S ⊆ Spec(A)containing x, y is then disconnected by S ∩Xf and S ∩Xe , so Spec(A) is totally disconnected.

12. Let A be an integral domain and M an A-module. An element x ∈M is a torsion element of M if Ann(x) 6= 0, that is ifx is killed by some non-zero element of A. Show that the torsion elements of M form a submodule of M . This submoduleis called the torsion submodule and is denoted by T (M ).

Let x, y ∈ T (M ) and c ∈ A. Then there are a, b 6= 0 in A such that ax = b y = 0. Since A is an integral domain,ab 6= 0 and we have ab (x + y) = b0+ a0= 0, so x + y ∈ T (M ). Also a(c x) = c0= 0, so c x ∈ T (M ).

If T (M ) = 0, the module M is said to be torsion-free. Show thati) If M is any A-module, then M/T (M ) is torsion-free.

Let x ∈ M/T (M ) and suppose a ∈ A\0 is such that ax = 0. Then any representative x of x in M is such thatax ∈ T (M ). But then there is b 6= 0 such that bax = 0 in M . Since ba 6= 0, we see x ∈ T (M ), so x = 0.

ii) If f : M →N is a module homomorphism, then f

T (M )

⊆ T (N ).Let x ∈ T (M ) and 0 6= a ∈A such that ax = 0. Then 0= f (ax) = a f (x), so f (x) ∈ T (N ).

iii) If 0→M ′→M →M ′′ is an exact sequence, then the sequence 0→ T (M ′)→ T (M )→ T (M ′′) is exact.Write M ′ f−→ M ′ g−→ M ′′. T ( f ) is injective because it is a restriction of the injective map f . T (g ) T ( f ) is zero

because it is a restriction of the zero map g f . If x ∈ T (M )∩ker(g ), then it is in im( f ), so there is y ∈M ′ such thatx = f (y). If 0 6= a ∈A is such that ax = 0, then f (ay) = 0; as f is injective, ay = 0, so y ∈ T (M ′).

iv) If M is any A-module, then T (M ) is the kernel of the mapping x 7→ 1⊗ x of M into K ⊗AM , where K is the field offractions of A.

48


Take S = A\0 and use (3.5), which gives an isomorphism K ⊗AM∼−→ S−1M taking (a/x)⊗m 7→ am/x. We

then have 1⊗m 7→ m/1= 0 in S−1M just if ([3.1]) there is x ∈ S such that x m = 0.2

13. Let S be a multiplicatively closed subset of an integral domain A. In the notation of Exercise 12, show that T (S−1M ) =S−1(T M ).

Let x ∈ T (M ) and 0 6= a ∈ Ann(x). Then for all s ∈ S we have a(x/s) = ax/s = 0 in S−1M , so S−1(T (M )) ⊆T (S−1M ). Conversely, suppose x/s ∈ T (S−1M ). Then there is a nonzero a/t ∈ S−1A such that ax/st = 0/1, so thereis u ∈ S such that uax = 0. But ua 6= 0 since A is an integral domain, so x ∈ T (M ) and x/s ∈ S−1(T (M )). ThusT (S−1M ) = S−1(T (M ))

Deduce that the following are equivalent:i) M is torsion-free.ii) Mp is torsion-free for all prime ideals p.iii) Mm is torsion-free for all maximal ideals m.

For each a ∈ A, the map la : x 7→ ax is an A-module homomorphism M → M ; it induces x/s 7→ ax/s in eachMp for p a prime. By (3.9), la is injective just if each localization (la)m for m maximal (or just prime) is injective. Bythe above, this is the same as demanding la/s is injective for all s ∈ S. But a module is torsion-free just if all la (andfriends) are injective, for a 6= 0.

14. Let M be an A-module and a an ideal of A. Suppose that Mm = 0 for all maximal ideals m⊇ a. Prove that M = aM .By (1.1), there is a bijective correspondence between maximal ideals m⊇ a and maximal ideals m′ of A/a. Now

if Mm = 0, then

0=Mm/(aM )m(3.4.iii)∼= (M/aM )m

(3.5)∼= Am⊗AM/aM

as Am-modules. For any x ∈ a, any s ∈A\m, and any m ∈M/aM we then have (x/s)⊗ m = (1/s)⊗ x m = 0 in Am⊗AM/aM , so this A/a-module is naturally isomorphic to (A/a)m ⊗A/aM/aM = 0. Now (M/aM )m/a ∼= (A/a)m/a ⊗A/aM/aM by (3.5), and [3.4] shows (A/a)m and (A/a)m/a are isomorphic, so we finally see each localization of M/aMat a maximal ideal of A/a is zero. Then (3.8) says that M/aM = 0. Thus M = aM .

15. Let A be a ring, and let F be the A-module An . Show that every set of n generators of F is a basis of F.Let ei be the standard basis of An and xi our generators; φ : ei 7→ xi is then a surjective homomorphism, and

⟨xi ⟩ will be a basis just if φ is also injective. By (3.9), φ is injective just if each φm is injective for m Ã A maximal.Thus without loss of generality we may assume A is local. Let N = ker(φ), so we have an exact sequence 0→N →F → F → 0. Tensoring with the residue field k = A/m gives an exact sequence k ⊗N → k ⊗ F → k ⊗ F → 0. Now(2.14.iii,iv) give k ⊗ F = k ⊗A⊕n ∼= (k ⊗A)⊕n ∼= kn . The map kn → kn is a surjection of vector spaces of the samedimension, hence an isomorphism, and thus k ⊗N = 0. Now [2.12] shows that N is finitely generated, and [2.2]gives k ⊗N = (A/m)⊗N ∼= N/mN = 0. Thus N = mN , and Nakayama’s Lemma (2.6) gives N = 0 (m being theJacobson radical of the local ring A). Thus φ is injective.

Deduce that every set of generators of F has at least n elements.Supposing m < n elements x1, . . . , xm generate F , then expanding this set at random by nonzero elements

y1, . . . , yn−m , we would have a set of n generators. By what we’ve proven above, this would be a basis. But sincethe xi are generators, we could write y1 =

∑mi=1 ai xi with not all ai = 0, contradicting this set being a basis.

We also showed this in [2.11]. (Surjections AmAn only occur for m ≥ n.)

2 Alternately, we may follow the book’s hint. Suppose x ∈ T (M ), and 0 6= a ∈ A is such that ax = 0. Then in K ⊗AM we have the equalities1⊗ x = a

a ⊗ x = 1a ⊗ ax = 1

a ⊗ 0= 0.For the other inclusion, we write K⊗AM as a direct limit. For each x ∈ S =A\0, we have a cyclic A-module Ax :=Ax ⊆K , and given x, y ∈ S,

we have natural inclusions Ax ,→Ax, y and Ay ,→Axy , given by a/x 7→ ay/xy and a/y 7→ ax/xy, so these modules Ax form a direct system, withan inclusion Ax ,→Ay just when x | y. Since every element ξ ∈K can be written as a/x for some a ∈A and x ∈ S, by [2.17], K = lim−→ Ax.

By [2.20], K⊗AM ∼= lim−→ (Ax⊗AM ). Now by [2.15], every element 1⊗m representing 0 is already equal to zero at some finite stage; that is, thereis some x ∈ S such that 1⊗m = 0 in Ax ⊗AM . But as an A-module Ax is isomorphic to A under a/x 7→ a, so we have a composite isomorphismAx ⊗AM →A⊗AM →M taking (a/x)⊗m 7→ a⊗m 7→ am. Since 0= 1⊗m = (x/x)⊗m 7→ xm we then have xm = 0, so x ∈ T (M ).

49


16. Let B be a flat A-algebra. Then the following conditions are equivalent:i) aec = a for all ideals a of A.ii) Spec(B)→ Spec(A) is surjective.iii) For every maximal ideal m of A we have me 6= (1).iv) If M is any non-zero A-module, then MB 6= 0.v) For every A-module M , the mapping x 7→ 1⊗ x of M into MB is injective.

B is said to be faithfully flat over A.Write f : A→ B for the map making B a flat A-algebra.i) =⇒ ii) : Recall that f ∗ : Spec(B)→ Spec(A) is given by q 7→ qc . Thus f ∗(pe ) = p for all p ∈ Spec(A), so f ∗ is

surjective.ii) =⇒ iii): Let m ∈ Max(A). If f ∗(n) = m, then we have f (m) ⊆ n, so me = B f (m) ⊆ n. If me = (1), then

f ∗(n) = f ∗

(1)

= (1) 6=m, so me 6= (1).iii) =⇒ iv): We use contraposition. Supposing iv) false, let M 6= 0 be an A-module such that MB = B ⊗AM = 0.

Since B is flat, inclusions M ′ ,→M induce injections M ′B MB , so M ′B = 0 for all submodules M ′ ⊆M . In particularthis is the case for all cyclic submodules Ax. Ax is isomorphic, as an A-module, to a quotient of A, say A/a. Now0= (Ax)B ∼= B⊗AA/a∼= B/aB = B/ f (a)B , using [2.2] and the definition a · b = f (a)b of the A-module structure onB , so ae = f (a)B = B . If m⊇ a is a maximal ideal of A, then also B =me , so iii) doesn’t hold.

iv) =⇒ v): We again use contraposition. Suppose v) is false, and let M be an A-module such that the canonicalmap to MB isn’t injective. Then there is a non-zero element x ∈M such that 1⊗x = 0 in MB . By flatness, the inclusionAx ,→ M induces an injection (Ax)B MB , but the image of this map is B ⊗ x = 0, so (Ax)B = 0 and iv) does nothold.

v) =⇒ i): We once again use contraposition. We always have a⊆ aec , by (1.17.i), so suppose aÃ A is such thata ( aec . Then the submodule aec/a of M := A/a is nonzero. By [2.2], MB = B ⊗A(A/a) ∼= B/aB = B/ae , so thenatural map M → MB is essentially the map A/a→ B/ae induced by f , whose kernel is f −1(ae )/a = aec/a, whichwe have noted is not zero.

17. Let A f−→ B g−→C be ring homomorphisms. If g f is flat and g is faithfully flat, then f is flat.Let j : N ,→ M be an inclusion of A-modules. g f is flat, so jC : NC MC is injective. Now

MC = C ⊗AM ∼= C ⊗B B ⊗AM ∼= (MB )C by (2.14.iv) and (2.15), and similarly for N . Since g isfaithfully flat, the canonical maps iN : NB → (NB )C and iM : MB → (MB )C are injective, and we havethe commutative diagram at right. fC is injective since g f is flat. If fB wasn’t an injection, iM fBwould not be injective. But iM fB = fC iN is a composition of injections.

NBfB //

iN

MBiM

(NB )C // fC

// (MB )C

18. Let f : A→ B be a flat homomorphism of rings, let q be a prime ideal of B and let p= qc . Then f ∗ : Spec(Bq)→ Spec(Ap)is surjective.

Since for all s ∈ S := A\p we have f (s) /∈ q, f induces a map ef : Ap→ Bq taking a/s 7→ f (a)/ f (s). We also havean A-algebra Bp = f (S)−1B , and since f (S)⊆ B\q=: T , [3.3] gives an isomorphism Bq = T −1B = U−1( f (S)−1B) =

U−1Bp, where U = t/1 ∈ Bp : t ∈ T . Now ef evidently factors through this ring: Ap → Bp → Bq. Since f isflat, (3.10) says the map Ap → Bp is flat. Since Bq is a localization of Bp, by (3.6) the map Bp → Bq is flat. Then[2.8.ii] shows that Bq is flat as an Ap-module. Now if p/s is an element of the maximal ideal pAp of Ap, we haveef (p/s) = f (p)/ f (s) ∈ f (p)Bq ⊆ qBq, so it is not the case that (pAp)

e = (1). Then by [3.16.iii], Bq is faithfully flatover Ap and the map f ∗ : Spec(Bq)→ Spec(Ap) is surjective

19. Let A be a ring, M an A-module. The support of M is defined to be the set Supp(M ) of prime ideals p of A such thatMp 6= 0. Prove the following results:i) M 6= 0 ⇐⇒ Supp(M ) 6=∅.

This follows from (3.8): M = 0 ⇐⇒ ∀p ∈ Spec(A) (Mp = 0).

ii) V (a) = Supp(A/a).Let p ∈ Spec(A) and S = A\p. By [3.8] and (3.4.iii), we have (A/a)p ∼= S−1A/S−1a. This is non-zero just if

(1/1) = S−1A 6= S−1a, so that 1/1 6= a/s for any a ∈ a and s ∈ S. By definition, this means there is no t ∈ S such thats t = at ∈ a∩ S. Redefining s ′ = st and a′ = at , without loss of generality t ′ = 1, so (A/a)p 6= 0 ⇐⇒ a∩ S =∅ Butsince S =A\p, we have a∩ S 6=∅ ⇐⇒ a⊆ p ⇐⇒ p ∈V (a).

50


iii) If 0→M ′→M →M ′′→ 0 is an exact sequence, then Supp(M ) = Supp(M ′)∪ Supp(M ′′).Let p ∈ Spec(A). By (3.3), localization gives an exact sequence 0 → M ′p → Mp → M ′′p → 0. If p ∈ Supp(M ′),

then M ′p 6= 0 injects into Mp, so p ∈ Supp(M ). If p ∈ Supp(M ′′), then Mp surjects onto M ′′p 6= 0, so p ∈ Supp(M ). Ifp /∈ Supp(M ′)∪ Supp(M ′′), then our exact sequence reduces to 0→ 0→Mp→ 0→ 0, so Mp = 0 and p /∈ Supp(M ).

iv) If M =∑

Mi , then Supp(M ) =⋃

Supp(Mi ).The inclusions Mi ,→ M give rise to injections (Mi )p Mp by (3.3). Thus if any (Mi )p 6= 0 we have Mp 6= 0,

so⋃

Supp(Mi ) ⊆ Supp(M ). On the other hand, we have a natural surjection⊕

Mi ∑

Mi = M , and (see nextparagraph) localization distributes over direct sums, so by exactness again we have a surjection

⊕

(Mi )pMp. Thusif all (Mi )p = 0, then Mp = 0.

To see localization distributes over direct sums, note

S−1

⊕

Mi

(3.5)∼= S−1A⊗A

⊕

Mi

[2.20]∼=⊕

(S−1A⊗Mi )(3.5)∼=

⊕

S−1Mi . (3.6)

Note that if we forgot tensor isn’t left exact (which I did, for a while), we would think we had counterexamplesto this. Consider Fp = Z/pZ and Q as Z-modules. Now Q is generated over Z by the elements 1

n , so it is a sum of

the submodules 1nZ. Thus the elements zn = 1⊗ 1

n generate Fp⊗ZQ over Fp . In fact all are zero, since the generator

zn = p⊗ 1pn = 0. We can view Fp ⊗ZQ as the direct limit of Mn = Fp ⊗Z

1nZ along the maps Mn→Mmn induced by

1mnZ ,→

1nZ, and this shows that the map Mn→M pn is zero. Thus Fp ⊗ZQ= 0 has empty support. Now the cyclic

Fp -modules Mn are all isomorphic to Fp , for we have isomorphisms Mn∼= 1

nZ/pnZ ∼= Z/pZ, which has support

(0). If tensor were left exact, the inclusion 1nZ ,→Q would induce an injection MnFp ⊗ZQ= 0, and we would

have (0) ⊆∅, which is obviously false. In fact the induced map is 1⊗ 1n 7→ zn = 0.

v) If M is finitely generated, then Supp(M ) =V (Ann(M )) (and is therefore a closed subset of Spec(A)).Let x1, . . . , xn be generators for M . Then Axi are cyclic modules, so there are ai ÃA such that the maps a 7→ axi

induce isomorphisms A/ai∼−→Axi . Since M =

∑

i Axi , by iv) and i) of this exercise, and [1.15.iv],

Supp(M ) =⋃

Supp(Axi ) =⋃

Supp(A/ai ) =⋃

V (ai ) =V

⋂

ai

.

Now a ∈A annihilates M just if for each i we have axi = 0, and this happens exactly when a ∈ ai , so⋂

ai =Ann(M )and Supp(M ) =V (Ann(M )).

vi) If M , N are finitely generated, then Supp(M ⊗AN ) = Supp(M )∩ Supp(N ).For any p ∈ Spec(A), (3.7) gives (M ⊗AN )p = Mp ⊗Ap

Np. Now the localizations Mp and Np are again finitelygenerated modules over the local ring Ap, so [2.3] shows the tensor product is non-zero if and only if both factorsare nonzero. So p ∈ Supp(M ⊗AN ) just if p is in both Supp(M ) and Supp(N ).

vii) If M is finitely generated and a is an ideal of A, then Supp(M/aM ) =V (a+Ann(M ));M/aM ∼= A/a⊗AM by (3.5), so by iv) we have Supp(M/aM ) = Supp(A/a)∩ Supp(M ). By ii) and v) above, and

[1.15.iii], this is V (a)∩V (Ann(M )) =V (a∪Ann(M )) =V (a+Ann(M )).

viii) If f : A→ B is a ring homomorphism and M is a finitely generated A-module, then Supp(B⊗AM ) = f ∗−1(Supp(M )).Let q ∈ Spec(B), and p= qc . We show q ∈ Supp(MB ) ⇐⇒ p ∈ Supp(M ).

(B ⊗AM )q(3.5)∼= Bq⊗AM

(2.14.iv)∼= (Bq⊗ApAp)⊗AM

(2.15)∼=(3.5)

Bq⊗ApMp.

Now if Mp = 0, clearly (MB )q = 0. Thus we have the containment Supp(B ⊗AM )⊆ ( f ∗)−11(Supp(M )).On the other hand, suppose 0 = (MB )q ∼= Bq ⊗Ap

Mp, and let x1, . . . , xn be generators of M over A. Then theequations (1/1)⊗ (xi/1) = 0 hold in Bq⊗Ap

Mp By (2.13) there is a finitely generated Ap-submodule Ni ⊆ B wherealready (1/1)⊗ (xi/1) = 0. Let N =

∑ni=1 Ni ⊆ B ; then N is finitely generated, and (1/1)⊗ (x/1) = 0 in N⊗Ap

Mp forall x ∈Mp, so N ⊗Ap

Mp = 0. But then by [2.3] we have either N = 0 or Mp = 0, and we have 1/1 ∈N , so Mp = 0.

51


The⊆ containment holds for any module M , but the⊇ doesn’t necessarily hold for non-finitely generated mod-ules. Here is an obvious counterexample, thanks to Angelo Vistoli3: Take f : A=ZZ/pZ= B , for p > 0 a primenumber, and M =Q. Then B⊗AM = 0 since 1⊗1/1= 1⊗ p/p = p ⊗ 1

p = 0⊗ 1p = 0, and so Supp(MB ) = Supp(0) =∅

by i). On the other hand, Supp(M ) = Spec(Z) since no nonzero element of Z annihilates an element ofQ.4 Now theonly nontrivial ideal of B =Z/pZ is (0), and (0)c = f −1

(0)

= (p), so ( f ∗)−1(Spec(Z)) = (0) 6=∅.

To see the delicacy of the assumption of finite generation, it is illuminating to look at a failed proof and see whatgoes wrong. First, suppose M =Ax ∼=A/a is cyclic. Then MB = B ⊗AM ∼= B/aB = B/ae by [2.2] and the definitionof the A-module structure on B . Thus Supp(MB ) =V (Ann(B/ae )) =V (ae ) by v), while Supp(M ) =V (Ann(A/a)) =V (a). By (1.17.i) ae ⊆ q =⇒ a⊆ aec ⊆ qc = p, while a⊆ p= qc =⇒ ae ⊆ qc e ⊆ q, so Supp(MB ) = ( f

∗)−1(Supp(M ))for M cyclic. Now suppose M is generated over A by some xi , so MB is generated over B by the 1 ⊗ xi . Writeai =AnnA(xi ). Now, using the cyclic case at the line break,

Supp(MB ) = Supp

∑

B(1⊗ xi )

iv)=⋃

Supp(B(1⊗ xi ))?=⋃

Supp(B ⊗AAxi )

=⋃

( f ∗)−1(Supp(Axi )) = ( f∗)−1

⋃

Supp(Axi )

iv)= ( f ∗)−1(Supp(M )).

What goes wrong is the step labeled with a question mark, for tensor is not left exact! The map B ⊗AAxi → B ⊗AMinduced by the inclusion Axi ,→M need not be injective, so the support of the submodule of MB generated by 1⊗ xican easily be smaller than Supp(B⊗AAxi ). This happens, for example, in our “counterexample” in iv), where M =Q,A=Z, and B = Fp , using that 0= Fp · (1⊗

1nZ) is not Fp ⊗Z

1nZ∼= Fp ⊗ZZ∼= Fp :

∅= SuppFp

∑

Fp · (1⊗ 1/n)

iv)=⋃

SuppFp(Fp · (1⊗ 1/n))⊆

⋃

SuppFp(Fp ⊗Z(1/n)Z) =

⋃

SuppFp(Fp ⊗ZZ)

=⋃

( f ∗)−1(SuppZ(Z)) = ( f∗)−1(Spec(Z)) = ( f ∗)−1((p)) = (0).

This point may not be that subtle, but it eluded me for longer than I care to admit.

20. Let f : A→ B be a ring homomorphism, f ∗ : Spec(B)→ Spec(A) the associated mapping. Show thati) Every prime ideal of A is a contracted ideal ⇐⇒ f ∗ is surjective.

By definition, f ∗ is surjective if and only if for every p ∈ Spec(A) there is q ∈ Spec(B) such that p= f ∗(q) = qc ,meaning every prime ideal of A is a contracted ideal.

ii) Every prime ideal of B is an extended ideal =⇒ f ∗ is injective.Suppose q1 = ae

1 and q2 = ae2 are prime ideals of B whose images under f ∗ are equal. That means, by the definition

of f ∗, that aec1 = qc

1 = qc2 = aec

2 . But then extending again, and using (1.17.ii), we get q1 = ae1 = aec e

1 = aec e2 = ae

2 = q2.

Is the converse of ii) true?No. Note that we always have qc e ⊆ q, so the trick will be to sabotage all extensions of primes so that they are

not themselves prime, and the containment is proper. Let A be a ring and consider the ring5 A[ε] := A[x]/(x2).Since there is a quotient map π : A[ε] A with kernel (ε), each prime p Ã A gives rise to a distinct prime π−1(p)of A[ε] containing (ε). Since ε2 = 0, all primes of A[ε] contain (ε), so these are all the primes of A[ε]. We canwrite them as p+ = π−1(p) = pA+ (ε). Now consider the inclusion j : A → A[ε]. Then for a prime p Ã A wehave pe = pA[ε] = pA+ pε. This ideal is not prime, since A[ε]/pe ∼= (A/p)[ε] is not a domain. The prime idealsp+ properly contain pe . Thus no extended ideal pe is prime, and no prime ideal p+ is extended. On the other handj ∗ : p+ 7→ p is a bijective function Spec(A[ε])↔ Spec(A).

3 http://mathoverflow.net/questions/50406/extension-of-scalars-and-support-of-a-non-finitely-generated-module4 In fact S−1Q∼=Q for any S =Z\q, by the map φ : S−1Z⊗ZQ∼= S−1Q→Q taking n

s ⊗ab 7→

nab s . To see injectivity, note that in S−1Z⊗ZQ

we have ns ⊗

ab =

1s ⊗

nab =

1s ⊗

s nasb =

11 ⊗

nasb , so any sum of decomposable elements ni

si⊗ ai

bican be written as 1⊗ z, where z =

∑

ini aisi bi

, and

then φ(1⊗ z) = z = 0 ⇐⇒ 1⊗ z = 0. φ is surjective since 1/1 ∈ S−1Z.5 of “dual numbers”: cf. http://en.wikipedia.org/wiki/Dual_numbers

52

http://mathoverflow.net/questions/50406/extension-of-scalars-and-support-of-a-non-finitely-generated-module

http://en.wikipedia.org/wiki/Dual_numbers


21. i) Let A be a ring, S a multiplicatively closed subset of A, and φ : A→ S−1A the canonical homomorphism. Show thatφ∗ : Spec(S−1A)→ Spec(A) is a homeomorphism of Spec(S−1A) onto its image in X = Spec(A). Let this image be denotedby S−1X.In particular, if f ∈A, the image of Spec(Af ) in X is the basic open set Xf (Chapter 1, Exercise 17).

The one-to-one correspondence given by (3.11.iv) is a bijection between Y = Spec(S−1A) and S−1X = p ∈X : p∩S =∅, given by contractionφ∗ and extension of prime ideals. In particular, pec = p for primes p⊆A\S.6 By [1.21.i],φ∗ is continuous. Since all ideals of S−1A are extended ideals by (3.11.i), to prove φ∗|S−1X is a homeomorphism ontoits image, it is enough to show φ∗ sends a non-empty closed set V (ae ) ⊆ Y to the closed subset V (a)∩ S−1X ⊆ X.Indeed, if a⊆ p⊆A\S, then ae ⊆ pe 6= (1), and if ae ⊆ pe 6= (1), then a⊆ aec ⊆ pec = p.

Now let S = 1, f , f 2, . . . for some f ∈ A. Then S−1p ∈ Spec(Af ) just if p∩ S = ∅, and since p is prime, thishappens just if f /∈ p, so that p ∈Xf .

ii) Let f : A→ B be a ring homomorphism. Let X = Spec(A) and Y = Spec(B), and let f ∗ : Y → X be the mappingassociated with f . Identifying Spec(S−1A) with its canonical image S−1X in X, and Spec(S−1B) (= Spec( f (S)−1B)) withits canonical image S−1Y in Y , show that S−1 f ∗ : Spec(S−1B)→ Spec(S−1A) is the restriction of f ∗ to S−1Y , and thatS−1Y = f ∗−1(S−1X ).

Let q ∈ Y . Then f (S) ∩ q 6= ∅ ⇐⇒ ∃s ∈ S

f (s) ∈ q

⇐⇒ ∃s ∈ S

s ∈ f −1(q)

⇐⇒ S ∩ f −1(q) 6= ∅, soq ∈ S−1Y ⇐⇒ f ∗(q) ∈ S−1X ⇐⇒ q ∈ ( f ∗)−1(S−1X ), showing S−1Y = ( f ∗)−1(S−1X ).

S−1q //_

(S−1 f )∗

q_

f ∗

S−1

f −1(q) // f −1(q)

Spec(S−1B) ≈

(φ∗f (S))−1

//

(S−1 f )∗

S−1Y //

f ∗

Spec(B)

f ∗

Spec(S−1A) ≈

(φ∗S )−1

// S−1X // Spec(A)

Now S−1 f : S−1A→ S−1B is given by as 7→

f (a)f (s) . Let q ∈ S−1Y and a/s ∈

S−1A. We have

as∈ (S−1 f )−1(S−1q) ⇐⇒ ∃q ∈ q ∃t ∈ S

f (a)f (s)

=q

f (t )

⇐⇒ ∃u, t ∈ S ∃q ∈ q [ f (u t a) = f (u s)q ∈ q].

This certainly is the case if f (a) = q ∈ q (take t = s and u = 1), and iff (a) /∈ q, then since f (S)∩ q = ∅ we have f (Sa)∩ q = ∅ (q being prime), soa/s /∈ (S−1 f )−1(S−1q). Thus (S−1 f )−1(S−1q) = S−1

f −1(q)

, giving the com-mutative diagrams at left.

iii) Let a be an ideal of A and let b= ae be its extension in B. Let f : A/a→ B/b be the homomorphism induced by f . IfSpec(A/a) is identified with its canonical image V (a) in X, and Spec(B/b) with its image V (b) in Y , show that f ∗ isthe restriction of f ∗ to V (b).

Let q ∈ V (b). Then a ⊆ aec = bc ⊆ qc = f ∗(q), so f ∗(V (b)) ⊆ V (a). Nowwrite π : A A/a and $ : B B/b. Then by definition, f

π(a)

= f (a) + b =$

f (a)

, so f π=$ f . Taking ∗’s and using [1.21.vi],π∗ f ∗ = f ∗$∗, givingthe commutative diagram at right.

Spec(B/b) $∗

≈//

f ∗

V (b) //

f ∗

Spec(B)

f ∗

Spec(A/a) π∗

≈// V (a)

// Spec(A)

iv) Let p be a prime ideal of A. Take S =A\p in ii) and then reduce mod S−1p as in iii). Deduce that the subspace f ∗−1(p)of Y is naturally homeomorphic to Spec(Bp/pBp) = Spec(k(p)⊗AB), where k(p) is the residue field of the local ring Ap.Spec(k(p)⊗A B) is called the fiber of f ∗ over p.

Making the identifications of ii) and iii), we have the commutative diagram at right.Now k(p) = Ap/pAp is a field, so its prime spectrum is (0). Following the definitionsof the inclusions, we see Spec(k(p)) corresponds to the set of primes in Spec(Ap) thatcontain pAp, which in turn corresponds to the set of primes in Spec(A) that are containedin p and contain p, or in other words the singleton p. Thus Spec(Bp/pBp) is identifiedwith the preimage ( f ∗)−1(p) of p. Now writing T = (A/p)\0 for the image

Spec(Ap/pAp) _

Spec(Bp/pBp) _

( fp)∗oo

Spec(Ap) _

Spec(Bp) _

( fp)∗

oo

Spec(A) Spec(B)f ∗

ooof S =A\p in A/p, we have

k(p)⊗AB = (Ap/pAp)⊗AB(3.4.iii)∼= (A/p)p⊗AB

[3.4]∼= T −1(A/p)⊗AB(3.5)∼= T −1(A/p)⊗A/pA/p⊗AB

[2.2]∼= T −1(A/p)⊗A/pB/pB(3.5)∼= T −1(B/pB)

[3.4]∼= S−1(B/pB)(3.4.iii)∼= Bp/pBp.

6 by (3.11.ii), pec =⋃

s∈S (p : s), and since S ∩ p=∅, we have sa ∈ p ⇐⇒ a ∈ p, and pec = p. This is closely related to (3.16).

53


22. Let A be a ring and p a prime ideal of A. Then the canonical image of Spec(Ap) in Spec(A) is equal to the intersection ofall the open neighborhoods of p in Spec(A).

Write S = A\p. The canonical image S−1X of Spec(Ap) in Spec(A) is the set of primes q Ã A that don’t meetA\p, or in other words are contained in p. By [1.17], the basic open sets Xf form a basis of Spec(A), so every openneighborhood of p contains some Xf , and the intersection of all open neighborhoods of p is Z :=

⋂

Xf : p ∈Xf .

Now if p ∈Xf , we have f /∈ p, so for any prime q⊆ p we a fortiori have f /∈ q; so every open set Xf 3 p we haveS−1X ⊆Xf . Thus S−1X ⊆ Z . On the other hand, suppose q /∈ S−1X. Then q meets S =A\p, so there exists f ∈ q\p.Then p ∈Xf but q /∈Xf , and therefore q /∈ Z . Thus Z ⊆ S−1X.

23. Let A be a ring, let X = Spec(A) and let U be a basic open set in X (i.e., U =Xf for some f ∈A: Chapter 1, Exercise 17).i) If U =Xf , show that the ring A(U ) =Af depends only on U and not on f .

Write S f = 1, f , f 2, . . ., and note that p ∈ Xf ⇐⇒ f /∈ p ⇐⇒ S f ∩ p = ∅, since p is prime. Recall ([3.7.ii])that the saturation S f is defined to be the complement of the union of all prime ideals which do not meet S f , so that

T := S f = Sg . Recall the natural homomorphisms ρTf : Af → T −1A and ρT

g : Ag → T −1A of [3.2], and that, by [3.8],

since T is the saturation of S f and of Sg , these maps are isomorphisms. Then ρgf

:= (ρTg )−1 ρT

f is an isomorphism

Af∼−→Ag , and we have the following commutative diagram:

Aφ f

""

φg

φT

Af∼

ρTf

// T −1A Ag .∼

ρTg

oo

Since φ f is an epimorphism (Eq. 3.1 in [3.3]) it follows from the commutativity of the two small triangles thatρg

f : Af → Ag is the unique isomorphism of the two rings making the big triangle commute: φg = ρgf φ f . Thus

54


A(U ) is unique up to a unique isomorphism.7

ii) Let U ′ = Xg be another basic open set such that U ′ ⊆ U . Show that there is an equation of the form g n = u f forsome integer n > 0 and some u ∈ A, and use this to define a homomorphism ρ : A(U ) → A(U ′) (i.e., Af → Ag ) bymapping a/ f m to au m/g mn . Show that ρ depends only on U and U ′. This homomorphism is called the restrictionhomomorphism.

If Xg ⊆ Xf , then by definition every prime p not containing g doesn’t contain f .8 By de Morgan’s laws, everyprime containing f contains g . Intersecting these sets of primes, by (1.14) we have g ∈ r

(g )

⊆ r

( f )

. Thus there

7 An alternate, more direct proof not using saturation is as follows.If Xf = Xg , by [1.17.iv], r

( f )

= r

(g )

. In particular, f ∈ r

(g )

and g ∈ r

( f )

, so there are u, v ∈ A and n, m > 0 such that g n = u f andf m = vg .

By (3.2), φg will induce a unique isomorphism ρ : Af →Ag such that φg = ρ φ f : A→Af →Ag just if

• For all n ≥ 0, we havef n

1a unit of Ag ;

• φg (c) =c1= 0 =⇒ ∃n ≥ 0 [ f n c = 0];

• Every element of Ag is of the formc1

f k

1

.

To prove the first item it will suffice to show f /1 is a unit in Ag . But u f = g n in A, so (u/g n)( f /1) = 1/1 in Ag , so that f /1 has inverse u/g n .Note that also (u/1)( f /g n), so u/1 is a unit of Ag as well.

For the second item, suppose a/1 = 0 in Ag . Then by definition there is k ≥ 0 such that g k a = 0 in A. Multiplying by vn g n−k we getf mnc = vnu f a = vn g na = 0.

For the third item, note that since f m = v g in A, we have ( f m/1)(1/g ) = v/1 in Ag . Multiplying by ( f /1)−m , we see 1/g = (v/1)( f /1)−m in

Ag . Thus an arbitrary element a/g l of Ag can be written as (av l /1)( f /1)−l m .

Yet another proof, this time not using universal properties or saturations, follows. I include it because I took the time to work it out, but thereader is advised to skip it.

Define ρgf : Af →Ag by c/ f k 7→ c( f ′)k and ρ f

g : Ag →Af by c/g k 7→ c(g ′)k , where f ′ = u/g n and g ′ = v/ f m Then

(ρ fg ρ

gf )

cf k

= ρ fg

cuk

g kn

=cukvkn

f k mn.

But cukvkn/ f k mn = c/ f k in Af , for

f kcukb kn = c(u f )k vkn = c g kn vkn = c(v g )kn = c f k mn

in A. Symmetrically ρgf ρ f

g = idAg. It remains to show these maps are well-defined homomorphisms. By symmetry, it will suffice to do so for

ρgf . Suppose c/ f k = d/ f l in Af . Then by definition there is p ≥ 0 such that f p f lc = f p f kd . For ρ

gf to be well-defined we require that

cuk

g kn= ρg

f

cf k

= ρgf

df l

=du l

g ln

in Ag . But this means, by definition, there is q ≥ 0 such that

c f l g q uk+l = g q (u f )lcuk = g q g lncuk = g q g kndu l = g q (u f )kdu l = d f k g q uk+l

in A. Take q = pn. Then, as hoped,

cf l g pn uk+l = c f l(u f )p uk+l = (c f l f p )uk+l+p = (d f k f p )uk+l+p = d f k(u f )p uk+l = d f k g pn uk+l .

By definition, ρgf is multiplicative, for we have

ρgf

cf k

ρgf

df l

= c( f ′)k d ( f ′)l = (cd )( f ′)k+l = ρgf

cf k· d

f l

.

Finally, for additivity, note that

ρgf

cf k+

df l

= ρgf

c f l + d f k

f k+l

= (c f l + d f k )( f ′)k+l =(c f l + d f k )uk+l

g n(k+l );

ρgf

cf k

+ρgf

df l

= c( f ′)k + d ( f ′)l =cuk

g nk+

du l

g nl=

cuk g nl + du l g nk

g n(k+l )=

cuk (u f )l + du l (uf )k

g n(k+l ).

8 From this, we see that the union of primes not meeting Sg = 1, g , g 2, . . . is a subset of the union of primes not meeting S f , so by the

definition of saturation in [3.7.ii], we have S f ⊆ Sg , so [3.3] and [3.8] give us a unique homomorphism S f−1A → Sg

−1A commuting with

55


is n > 0 so that g n ∈ ( f ). Then for some u ∈A, g n = u f . To define a unique homomorphism ρ : Af →Ag such thatφg = ρ φ f : A→ Af → Ag , by (3.1) it suffices to show that φg ( f ) = f /1 (and hence φg ( f

n)) is a unit. But sinceg n = u f in A we have ( f /1)(u/g n) = 1 in Ag . Thus a unique ρg

f : Af →Ag exists such that φg = ρgf φ f and it must

take 1/ f 7→ u/g n so that ( f /1)ρ(1/ f ) = 1/1.To show uniqueness, suppose ρ f

f ′ : Af ′∼−→Af and ρg ′

g : Ag∼−→Ag ′ are canonical isomor-

phisms, commuting with the canonical maps φ from A, and let ρgf and ρg ′

f ′ be the uniquemaps given in the preceding paragraph. Canonicity means that all the triangles in the dia-gram at right commute. That “ρ depends only on U and U ′” can mean nothing strongerthan that the outer square commutes. Now

ρg ′

f ′ φ f ′ =φg ′ = ρg ′g φg = ρ

g ′g ρ

gf φ f = ρ

g ′g ρ

gf ρ

ff ′ φ f ′ .

Af

ρgf // Ag

ρg ′g∼

Aφ f

__φg

??

φ f ′

φg ′

Af ′

ρ ff ′∼

OO

ρg ′

f ′

// Ag ′

As φ f ′ is an epimorphism (Eq. 3.1 in [3.3] again), ρg ′

f ′ = ρg ′g ρ

gf ρ

ff ′.

iii) If U =U ′, then ρ is the identity map.ρU

U = idA(U ) satisfies the equation φU = ρUU φU , and is unique since φU is an epimorphism by Eq. 3.1 of [3.3].

iv) If U ⊇U ′ ⊇U ′′ are basic open sets in X, show that the diagram

A(U ) //

$$

A(U ′′)

A(U ′)

99

(in which the arrows are restriction homomorphisms) is commutative.

S−1A

ρTS !!

ρVS

T −1A

ρVT !!

A

φS

OO

φT

==

φV

// V −1A

Recall that [3.3] and [3.8] give us, for multiplicatively submonoids S ⊆ T ⊆ Aunique homomorphisms ρT

S : S−1A→ T −1A such that φT = ρTS φS , where φS : A→

S−1A is the epimorphic (Eq. 3.1) canonical map. If S ⊆ T ⊆ V ⊆ A are multiplicativesubmonoids, we have

φV = ρVT φT = ρ

VT ρ

TS φS ,

but ρVS φS =φV as well, so sinceφS is epimorphic, ρV

T ρTS = ρ

VS . Now let U =X f ⊇

U ′ =X f ′ ⊇U ′′ =X f ′′ and S = S f ⊆ T = S f ′ ⊆V = S f ′′ , and use ii).

v) Let x (= p) be a point of X. Show thatlim−→U3x

A(U )∼=Ap.

Let, again, S =A\p. For f , g ∈ S, write f ≤ g if r

(g )

⊆ r

( f )

, or equivalently if Xg ⊆Xf . This makes S intoa directed pre-order, for

• r

( f )

⊆ r

( f )

=⇒ f ≤ f ,

• f ≤ g ≤ h implies r

(h)

⊆ r

(g )

⊆ r

( f )

, which implies f ≤ h, and

• if f , g ∈ S we have f g ∈ S, and r

( f g )

⊆ r

( f )

∩ r (

g )

.

Now one can take direct limits over directed pre-orders, but the result is the same if we collapse the pre-order to itsskeleton, or arbitrarily take one element in each equivalence class, where f ≡ g ⇐⇒ f ≤ g ≤ f , and then take thedirect limit. Since we defined direct limits over partial orders, let’s do that. For each basic open set U , take just onerepresentative f such that U = Xf . The uniqueness from part ii) makes our choice, up to a unique isomorphism,irrelevant. Then the restricted pre-order (S ′, ≤) is a directed set. Consider the system B= (Af , ρg

f ) f , g∈S ′ , where therestriction map ρg

f of ii) is defined if f ≤ g . Now S ′ is a directed set and parts iii) and iv) of this exercise give axioms

the maps φ from A. Recall that [3.8] gives us unique isomorphisms Af∼−→ S f

−1A and Sg−1A

∼−→ Ag commuting with the φ; letting ρgf

be the

composition Af∼−→ S f

−1A→ Sg−1A

∼−→Ag , we have a unique ring homomorphism Af →Ag such that φg = ρgfφ f . But this doesn’t give us an

explicit expression for the homomorphism.

56

Chapter 3: Rings and Modules of Fractions Ex. 3.23 12 *

(1) and (2) of [2.14] defining compatibility conditions for maps in a direct system. Thus B is a direct system. WriteB = lim−→B for the limit, and ρ f : Af → B for the canonical map.

Now for each f ∈ S ′ we have S f = 1, f , f 2, . . . ⊆ S, so by [3.3] and [3.8] there is a unique map σ f := ρSf : Af =

S−1f A→ S−1A=Ap such that σ f φ f =φS . If f ≤ g ∈ S ′, then

σg ρgf φ f = ρ

Sg ρ

gf φ f = ρ

Sf φ f = σ f φ f ,

where the middle equality comes from the slightly generalized version of part iv), and then since φ f is an epimor-phism (Eq. 3.1 of [3.3]), σg ρ

gf = σ f . Since the σ f meet this compatibility condition, by [2.16], there then exists a

unique ring map σ : B→Ap such that σ ρ f = σ f for all f ∈ S ′. It remains to show σ is a bijection.Let b ∈ B be such that σ(b ) = 0 in Ap. By [2.15] there is some f ∈ S ′ such that b has a representative in Af , so let

a ∈ A and n ≥ 0 be such that ρ f (a/ f n) = b . Then we have 0= σ(b ) = σ(ρ f (a/ f n)) = σ f (a/ f n) = ρSf (a/ f n). That

means that considered as an element of Ap we have a/ f n = 0, so by [3.1] there is s ∈A\p= S such that sa = 0 in A.

But then in As f we have ρs ff (a/ f n) = as n/(s f )n = 0, so

b = ρ f (a/ f n) = ρs f (ρs ff (a/ f n)) = ρs f (0) = 0,

showing σ is injective.Let any element a/s ∈ S−1A= Ap be given. Then a/s ∈ As is of course such that ρS

s (a/s) = a/s , and if we letb = ρs (a/s), then

σ(b ) = σ(ρs (a/s)) = σs (a/s) = ρSs (a/s) = a/s .

Therefore σ is surjective, completing the proof.

The assignment of the ring A(U ) to each basic open set U of X, and the restriction homomorphisms ρ, satisfying theconditions iii) and iv) above, constitutes a presheaf of rings on the basis of open sets (Xf ) f∈A. v) and iv) says that the stalkof this presheaf at x ∈X is the corresponding local ring Ap.

23 12 *. Complete the description of a presheaf structure on X = Spec(A), by defining A(U ) for all open subsets U ⊆ X, not just

basic ones.Let V ⊆ X be an arbitrary open set, and let X fi

i∈I be all the basic open sets contained in V . Let Si be thesaturation of 1, fi , f 2

i , . . ., so that Si = A \⋃

Ui by [3.7.ii]. By [3.8] we have a canonical isomorphism A(Ui ) ∼=S−1

i A. Then SV =⋂

i∈I S j is a saturated set since it contains 1, and

xy ∈ SV ⇐⇒ ∀i ∈ I (xy ∈ Si ) ⇐⇒ ∀i ∈ I (x ∈ Si & y ∈ Si ) ⇐⇒ x ∈ SV & y ∈ SV .

NowSV =

⋂

i∈I

A\⋃

p∈Ui

p

=A\⋃

i∈I

⋃

p∈Ui

p=A\⋃

n

p : p ∈⋃

i∈I

Ui

o

=A\⋃

p∈V

p. (3.7)

Define A(V ) := S−1V A. If W ⊆ V is a smaller open set, Eq. 3.7 shows that SV ⊆ SW , so [3.3] and [3.8] give us a

natural restriction map ρWV : A(V )→ A(W ). [23.iii,iv] show that these maps define a presheaf, and [3.8] shows this

definition agrees with the other one if V is a basic open set. Since every open set contains a basic open set, the ringsA(Uf ) are cofinal in the A(V ), so the direct limits Ap

∼= lim−→V3pA(V )∼= lim−→Uf 3p

A(Uf ) are isomorphic and stalks areunchanged.

24. Show that the presheaf of Exercise 23 has the following property. Let (Ui )i∈I be a covering of X by basic open sets. For eachi ∈ I let si ∈ A(Ui ) be such that, for each pair of indices i , j , the images of si and s j in A(Ui ∩Uj ) are equal. Then thereexists a unique s ∈ A (= A(X )) whose image in A(Ui ) is si , for all i ∈ I . (This essentially implies that the presheaf is asheaf.)

Let Ui =X fi. By [1.17.v], X is compact , so there are finitely many U1, . . . , Un covering X. Thus

X \n⋂

i=1

V ( fi ) =n⋃

i=1

X \V ( fi )

=n⋃

i=1

X fi=X .

57


By [1.15.iii,i], ∅ =⋂n

i=1 V ( fi ) =

∑ni=1( fi )

, so no maximal ideal contains∑n

i=1( fi ), which then must be (1). Fix

m ≥ 1. (1.16) shows that for any ideals a, b Ã A, if a+ b = (1), then am + b = (1), since a ⊆ r (am). Applying thisrepeatedly to the sum

∑ni=1( fi ) = (1), taking a= ( f1), then ( f2), etc., we see

∑ni=1( f

mi ) = 1, so there are ai ∈ A such

thatn∑

i=1

ai f mi = 1. (3.8)

We first show uniqueness of s ∈A, assuming it exists. If s and s ′ both meet the conditions, then ρUiX (s) = ρ

UiX (s

′)for each canonical restriction map ρUi

X = φi : A→ A(Ui ), so ρUiX (s − s ′) = 0. We want to show t = s − s ′ is zero.

Now t ∈ ker(ρUiX ) if and only if, by [3.1], for some number mi ≥ 0 we have f mi

i t = 0. Let m = maxni=1 mi , so for

i = 1, . . . , n we have f mi t = 0. Multiplying t by the expression Eq. 3.8 for 1, we get

t = 1t =n∑

i=1

ai f m t =n∑

i=1

ai · 0= 0.

Now we must show existence. We initially restrict attention to the open cover U1, . . . , Un. Suppose si ∈ Afi

is given by b ′i / f mii . Let m = maxn

i=1 mi . Then setting bi = b ′i f m−mii , we have si = (b

′i / f mi

i )( fi/ fi )m−mi = bi/ f m

i .Write gi = f m

i .9 Now X fi∩X f j

=X fi f jby [1.17.i], and the images bi/gi = b j/g j in Afi f j

just if there is mi j ≥ 0 suchthat (gi g j )

mi j g j bi = (gi g j )mi j g m

i b j . (If X fi f j= ∅, then fi f j is nilpotent, by [1.17.ii], and then 1/1 = ( fi f j/ fi f j ) is

nilpotent, so 1= 0 ∈Afi giand the restrictions agree trivially.) Take p =maxi , j mi j ; then

g pi g p+1

j bi = g p+1i g p

j b j . (3.9)

We want to find an s such that s/1= bi/gi in each Afi, meaning there exists k such that g k+1

i s = g ki (s gi ) = g k

i bi inA. In an attempt to make this work and find s , fix j , take Eq. 3.8, with m to be determined later, and multiply bothsides by g k

j b j . Then

g kj b j =

n∑

i=1

ai b j g kj g m

i .

We want to use Eq. 3.9 to get one more g j on the right-hand side, so we should require m ≥ p+ 1 and k ≥ p. Then

g kj b j =

n∑

i=1

ai (b j g pj g p+1

i )g k−pj g m−p−1

iEq. 3.9=

n∑

i=1

ai (bi g p+1j g p

i )gk−pj g m−p−1

i = g k+1j

n∑

i=1

ai bi g m−1i .

In particular, k = p and m = p+1 work. Then if we take s =∑n

i=1 ai bi g pi , we have g p

j b j = g p+1j s for all j = 1, . . . , n,

so this s satisfies ρUj

X (s) = s j , as hoped.Now let Ui =V =Xh be an arbitrary element of the initial cover (hence not necessarily one of U1, . . . , Un). We

must show ρVX = si =: t . Now W j =Uj ∩V , for j = 1, . . . , n, are by [1.17.i] also basic open sets, and by assumption

ρW j

V (t ) = ρW j

Uj(s j ) = ρ

W j

X (s)[3.23.iv]= (ρ

W j

V ρVX )(s).

Thus ρW j

V (t −ρVX (s)) = 0 for j = 1, . . . , n. But the W j cover Xh = Spec(Ah ). Now by the uniqueness clause above,

applied to Ah instead of Aand the open cover W j instead of Uj , and implicitly using the isomorphisms (Ah ) f j /1∼=Ah f j

,

we have t = ρVX (s).

9 Though a posteriori it seems natural, the substitution of gi for fi and the conditions on k , m later I learned from looking over JinhyunPark’s solution: http://mathsci.kaist.ac.kr/~jinhyun/sol2/comm.html.

58

http://mathsci.kaist.ac.kr/~jinhyun/sol2/comm.html


25. Let f : A→ B, g : A→C be ring homomorphisms and let h : A→ B ⊗AC be defined by h(x) = f (x)⊗ g (x). Let X, Y ,Z, T be the prime spectra of A, B, C , B ⊗AC respectively. Then h∗(T ) = f ∗(Y )∩ g ∗(Z).

The first thing is to note that, as on p. 31, the suggested map h is not an A-algebra homomorphism. See [2.23].Instead define h : a 7→ f (a)⊗ 1= 1⊗ g (a).

That out of the way, let p ∈X and write k =Ap/pAp. We have

p ∈ h∗(T ) ⇐⇒ ∅ 6= (h∗)−1(p)[3.21.iv]≈ Spec(k ⊗AB ⊗AC )

(2.14.i,ii)≈ Spec(B ⊗Ak ⊗AC )

(2.14.i)≈ Spec(B ⊗A(k ⊗k k)⊗AC )

(2.15)≈ Spec

(B ⊗Ak)⊗k (k ⊗AC )

Now the only ring with empty spectrum is the zero ring, and a tensor product of vector spaces is nonzero just ifeach factor is nonzero. Thus p ∈ h∗(T ) just if k⊗AB and k⊗AC are nonzero. By [3.21.iv] again, this happens just ifp ∈ f ∗(Y ) and p ∈ g ∗(Z).

26. Let (Bα, gαβ) be a direct system of rings and B the direct limit. For each α, let fα : A→ Bα be a ring homomorphism suchthat gαβ fα = fβ whenever α≤β (i.e. the Bα form a direct system of A-algebras). The fα induce f : A→ B. Show that

f ∗

Spec(B)

=⋂

α

f ∗α

Spec(Bα)

.

To see the map f is well defined, choose any α and define f (a) = gα( fα(a)), where gα : Bα → B is the canonicalmap in the definition of the direct limit. Suppose γ ≥ α. Then

gα fα[2.14]= gγ gαγ fα = gγ fγ

Thus, for any α, β, find γ ≥ α, β; then gα fα = gγ fγ = gβ fβ, so the definition is independent of α.Let p ∈ Spec(A) be given, and set k = Ap/pAp. Recall from [2.20] that lim−→(Bα ⊗Ak) ∼= B ⊗Ak, so the two have

homeomorphic spectra. Now by [3.21.iv], p ∈ f ∗

Spec(B)

just if Spec(B⊗Ak) 6=∅ just if k⊗AB 6= 0. On the otherhand p ∈ f ∗α

Spec(Bα)

just if k ⊗ABα 6= 0. Now by [2.21], k ⊗AB = 0 just if for some α, k ⊗ABα = 0.

27. i) Let fα : A→ Bα be any family of A-algebras and let f : A→ B be their tensor product over A (Chapter 2, Exercise 23).Then

f ∗

Spec(B)

=⋂

α

f ∗α

Spec(Bα)

.

Recall from [2.23] that B = lim−→BJ , where each J is a finite set of α’s, BJ =⊗A

α∈J Bα, and if J = α1, . . . , αm andI = J ∪αm+1, . . . , αnwe define gJ I (b1⊗· · ·⊗bn) = b1⊗· · ·⊗bn⊗1⊗· · ·⊗1. For each J , fJ : a 7→ fα1

(a)⊗1⊗· · ·⊗1=a(1⊗· · ·⊗1) defines an A-algebra homomorphism A→ BJ independent of the apparent choice of non-1 coordinate bythe definition of the tensor product. Now we are in the situation of [3.26], and f ∗

Spec(B)

=⋂

J f ∗J

Spec(BJ )

. Butby iterated application of [3.25], we have f ∗J

Spec(BJ )

=⋂

α∈J f ∗α

Spec(Bα)

, so f ∗

Spec(B)

=⋂

J f ∗J

Spec(BJ )

.(To consider only singletons J = α is strictly speaking insufficient since without [3.25], we don’t know for I 3 αwhat relation exists between f ∗I

Spec(BI )

and f ∗α

Spec(Bα)

.)

ii) Let fα : A→ Bα be any finite family of A-algebras and let B =∏

α Bα. Define f : A→ B by f (x) = ( fα(x)). Thenf ∗

Spec(B)

=⋃

α f ∗α

Spec(Bα)

.Recall [1.22]: if eα has α-coordinate 1 and other coordinates 0, and bÃ B , then b=

∑

beα, and, writing πα : BBα for the canonical projection, B/b∼=

∏

Bα/πα(b), so pÃ B is prime just if it is of the form p+α = pαeα+∑

β 6=α(eβ)for some α and some prime pα Ã Bα. Recall that this gives a homeomorphism between Spec(B) and the disjointunion of the Spec(Bα). Now

f −1(p+α) = f −1b = (bα) ∈ B : ∃a ∈A∀β ( fβ(a) = bβ & fα(a) = bα ∈ pα)

= a ∈A : fα(a) ∈ pα= f −1α (pα), so

f ∗

Spec(B)

=⋃

α

f −1(p+α) : pα ∈ Spec(Bα)=⋃

α

f −1α (pα) : pα ∈ Spec(Bα)=

⋃

α

f ∗α

Spec(Bα)

.

59


iii) Hence the subsets of X = Spec(A) of the form f ∗

Spec(B)

, where f : A→ B is a ring homomorphism, satisfy theaxioms for closed sets in a topological space. The associated topology is the constructible topology on X. It is finer than theZariski topology (i.e., there are more open sets, or equivalently more closed sets).

For a homomorphism f : A→ B , write C f = f ∗

Spec(B)

for the closed set of the constructible topology definedby f . To finish checking the C f define a topology, we should ensure that∅ and X are among them. But X =CidA

, and∅= Cz for z : A→ 0, which has no prime ideals. Every V (a) closed in the Zariski topology is C f for the canonicalmap f : A A/a, so the constructible topology is at least as fine as the Zariski topology. It is not always strictlyfiner, as witness the zero ring or a ring with only one prime ideal.

iv) Let XC denote the set X endowed with the constructible topology. Show that XC is compact.To show every finite open cover of X has a finite subcover is the same as showing that for every collection of

closed sets of X with empty intersection, some finite subset has empty intersection.So let C f : f : A→ B f have empty intersection. Write g : A→ B =

⊗

f B f for the canonical map given by [2.23]making B an A-algebra. Then by [3.27.i],

∅=⋂

C f =⋂

f ∗

Spec(B f )

= g ∗

Spec(B)

,

so Spec(B) is empty. Since by (1.4) every ring where 0 6= 1 has a maximal ideal, B = lim−→JBJ = 0. But then by [2.21],

we have some BJ =⊗A

f∈J B f = 0 for some finite subset J of the f . Write h : A→ BJ . Applying [3.27.i] again, we have

∅= h∗

Spec(BJ )

=⋂

f∈J

f ∗

Spec(B f )

=⋂

f∈J

C f .

28. (Continuation of Exercise 27.)i) For each g ∈A, the set Xg (Chapter 1, Exercise 17) is both open and closed in the constructible topology.

If f : A→Ag is the canonical map, then C f is by (3.11.iv) the set of primes not meeting 1, g , g 2, . . ., so C f =Xgis closed in the constructible topology. On the other hand since the constructible topology is at least as fine as theZariski topology, Xg is open. More explicitly, let a= r

(g )

be the intersection of all primes containing g . Then forall primes p ∈ Spec(A)we have p ∈V (a) ⇐⇒ g ∈ p ⇐⇒ p /∈Xg , so Xg =X\V (a). But V (a) is C f for the canonicalmap f : A→A/a.

ii) Let C ′ denote the smallest topology on X for which the sets Xg are both open and closed, and let XC ′ denote the set Xendowed with this topology. Show that XC ′ is Hausdorff.

Let x, y ∈ XC ′ be distinct points. Then the corresponding primes px , py ∈ Spec(A) are not equal, so one is notstrictly contained in the other. Suppose without loss of generality that px 6⊆ py . Then there is f ∈ px \py , meaningy ∈Xf , but x ∈XC ′\Xf . But these are disjoint open sets by the definition of C ′.

iii) Deduce that the identity mapping XC →XC ′ is a homeomorphism. Hence a subset E of X is of the form f ∗

Spec(B)

for some f : A→ B if and only if it is closed in the topology C ′.First, the identity mapping XC →XC ′ is of course a bijection.Second, the topology C ′ is the topology generated by subbasic closed sets Xg and X \Xg for each g ∈A. (Thus a

closed set K of XC ′ is one that can be written as finite a union of arbitrary intersections of these.) But these subbasicsets are also closed in XC , so every closed set of XC ′ is closed in XC and the identity map is continuous.

Third, a continuous bijection ψ : Y ↔ Z between a compact space Y and a Hausdorff space Z is well known tobe a homeomorphism.10

iv) The topological space XC is compact, Hausdorff and totally disconnected.Since XC is compact, XC ′ is Hausdorff, and we have just shown XC ≈ XC ′ , it remains to see XC is totally dis-

connected. But this follows from our proof it is Hausdorff, for given any distinct x, y ∈ S ⊆ XC we have founddisjoint closed open sets U = Xf and V = XC \Xf separating them, whose union is the whole space XC , and soS = (U ∩ S)q (V ∩ S) is a disjoint union of closed open sets (in the relative topology) separating x from y.

10 It suffices to prove it takes open sets to open sets. Let U ⊆ Y be open; then its complement K = Y \U is closed. As a closed subsetof a compact space, K is compact. Since ψ is continuous, ψ(K) is compact. As a compact subset of a Hausdorff space, ψ(K) is closed. ThusZ\ψ(K) =ψ(Y \K) =ψ(U ) is open.

60


29. Let f : A→ B be a ring homomorphism. Show that f ∗ : Spec(B)→ Spec(A) is a continuous closed mapping (i.e., mapsclosed sets to closed sets) for the constructible topology.

Let any constructible closed set K ⊆ Spec(B) be given, and let g : B→C be such that K = g ∗

Spec(C )

. Then

f ∗(K) = f ∗

g ∗

Spec(C )

[1.21.vi]= (g f )∗

Spec(C )

is by definition a closed set of Spec(A) in the constructible topology.

30. Show that the Zariski topology and the constructible topology on Spec(A) are the same if and only if A/N is absolutelyflat (where N is the nilradical of A).

First suppose A/N is absolutely flat. Then by [3.11], X = Spec(A) in the Zariski topology is Hausdorff. Since theconstructible topology is at least as fine as the Zariski topology, the identity map XC →X is a continuous bijectionfrom a compact space to a Hausdorff space, hence a homeomorphism; see the footnote to [3.28.iii].

Now suppose X =XC . Then X is Hausdorff, so by [3.11], A/N is absolutely flat.

61

Chapter 4: Primary Decomposition

Proposition 4.8. Let S be a multiplicatively closed subset of A, and let q be a p-primary ideal.ii) If S ∩ p=∅, then S−1q is S−1p-primary, and its contraction in A is q.

The book states that “The verification that S−1q is primary is straightforward.” We complete the this verification.Suppose x/s , y/t ∈ S−1A are such that xy/s t ∈ S−1q. Then there are some z ∈ q and u ∈ S such that xy/s t = z/uin S−1A. That means there is v ∈ S such that v u xy = v s t z ∈ q ⊆ A. Since q is primary, either v u x ∈ q or yn ∈ qfor some n > 0. If v u x ∈ q, then v u x/s v u = x/s ∈ S−1q. If yn ∈ q, then (y/t )n = (1/t n)(yn/1) ∈ S−1q. Thusx/s ∈ S−1q or (y/t )n ∈ S−1q, showing S−1q is primary.

Proposition 4.12*.1 Let A be a ring, S a multiplicatively closed subset of A. WriteφS : A→ S−1A for the canonical map.For any ideal a, let S(a) denote the contraction along φS of S−1a in A (bottom of p. 53, [4.11]). The ideal S(a) is calledthe saturation of a with respect to S.i)⋃

s∈S (a : s) = x ∈A : ∃s ∈ S (s x ∈ a)= S(a) = aec ⊇ a.ii) S(0) = ker(φS ).iii) Let Sp =A\p for p a prime ideal of A. If q is p-primary, then Sp(q) = q.iv) Sp(0) is contained in every p-primary ideal of A.v) If S1 ⊆ S2 ⊆A are multiplicative submonoids, then S1(a)⊆ S2(a).vi) If bÃA is an ideal containing a, then S(a)⊆ S(b).

i): a ⊆ aec is part of (1.17.i). Since S−1a = S−1A · a = ae , by definition we have S(a) =

S−1ac = aec . Now

x ∈ S(a) ⇐⇒ x/1 ∈ S−1a ⇐⇒ ∃a ∈ A ∃s ∈ S (x/1 = a/s). By the definition of S−1A, this happens if and only ifthere exist t , s ∈ S and a ∈ a such that t s x = t a ∈ S · a= a. Thus x ∈ S(a) ⇐⇒ S x ∩ a 6=∅ ⇐⇒ ∃s ∈ S (s x ∈ a).This happens just if x ∈

⋃

s∈S (a : s).ii): (0)e = S−1(0) = (0), so S

(0)

= (0)ec = (0)c =φ−1S

(0)

= ker(φS ).iii): x ∈ Sp(q) ⇐⇒ ∃s ∈ Sp (s x ∈ q). Since s /∈ p= r (q) and q is primary, x ∈ q, so q⊆ qec = Sp(q)⊆ q.iv): Let q be p-primary. If x ∈ Sp(0), there is s ∈ Sp such that s x = 0 ∈ q. Since by definition s /∈ p = r (q), no

power of s is in q, so since q is primary, x ∈ q.v): If ∃s ∈ S1 (s x ∈ a), then s ∈ S2, so ∃s ∈ S2 (s x ∈ a). By i), we have S1(a)⊆ S2(a).vi): Apply (1.17.iv*) twice: a⊆ b =⇒ ae ⊆ be =⇒ S(a) = aec ⊆ bec = S(b).

EXERCISES1. If an ideal a has a primary decomposition, then Spec(A/a) has only finitely many irreducible components.

Let a =⋂n

i=1 qi be the primary decomposition. Recall from [1.20.iv] that the irreducible components of X =Spec(A/a) are the closed sets V (p), where p is a minimal prime ideal of A/a. These are of the form p= p/a for primespÃ A minimal over a, hence in bijection with the set of primes minimal over a. By (4.6), this is the set of minimalelements of the finite set of r (qi ), hence finite.

2. If a= r (a), then a has no embedded prime ideals.a= r (a) =

⋂

p ∈ Spec(A) : a⊆ p gives a “primary decomposition” of a since prime ideals are primary (clearly,p. 50). The scare quotes are because there may be infinitely many primes. Now if we take the intersection over onlythe minimal primes over a, the intersection is clearly still a, and by construction there are no embedded primes. Theintersection still needn’t be finite however. Take for example an infinite direct product A=

∏

i∈I ki of fields, and foreach i ∈ I let pi = 0×

∏

j 6=i k j . Then A/pi∼= ki , is a field, so pi is prime, and the zero ideal (0) is the intersection

of the pi , but if any of these is omitted, the intersection contains non-zero elements.

1 This is not a proposition from the book. It is original, and designed to be used in and generalize some parts of [4.10–13].

62

Chapter 4: Primary Decomposition Ex. 4.3

3. If A is absolutely flat, every primary ideal is maximal.Let qÃA be primary, and p= r (q). By [2.28], A/q is absolutely flat, and every zero-divisor is nilpotent. But by

[2.28] again, every non-unit is a zero-divisor, so every non-unit is nilpotent. By [1.10], it follows that A/q has exactlyone prime ideal; hence is a local, absolutely flat ring. But then by [2.28] yet again, A/q is a field, and it follows thatq was maximal.

4. In the polynomial ring Z[t ], the ideal m= (2, t ) is maximal and the ideal q= (4, t ) is m-primary, but is not a power ofm.Z[t ]/m∼=Z/(2) is a field, while Z[t ]/q∼=Z/(4) has all zero-divisors (0 and 2) nilpotent, and so is primary. The

radical r (q) = r

(4)+(t )

= r

r (4)+ r (t )

= r (2, t ) =m by (1.13.v,vi). The powers of m are (2, t ), m2 = (4, 2t , t 2),etc. q is contained in m, properly since 2 /∈ q, and contains m2, again properly since t /∈ m2. For n ≥ 2 we havemn ⊆m2 ( q(m, so q is not a power of m.

5. In the polynomial ring K[x, y, z] where K is a field and x, y, z are independent indeterminates, let p1 = (x, y),p2 = (x, z), m= (x, y, z); p1 and p2 are prime, and m is maximal. Let a= p1p2. Show that a= p1∩p2∩m2 is a reducedprimary decomposition of a. Which components are isolated and which are embedded?

Write A=K[x, y, z]. Now A/p1∼=K[z] and A/p2

∼=K[y] are integral domains, and A/m∼=K is a field. We havethe following five equations:

a= p1p2 = (x, y)(x, z) = (x2, xy, x z, y z);

m2 = (x, y, z)(x, y, z) = (x2, y2, z2, y z, x z, xy);

p1 ∩m2 = (x, y)∩ (x2, y2, z2, y z, x z, xy) = (x2, y2, x z, y z, xy);

p2 ∩m2 = (x, z)∩ (x2, y2, z2, y z, x z, xy) = (x2, z2, y z, x z, xy);

p1 ∩ p2 = (x, y)∩ (x, z) = (x, y z);

so none of the last three pairwise intersections is a. On the other hand,

p1 ∩ (p2 ∩m2) = (x, y)∩ (x2, z2, y z, x z, xy) = (x2, xy, y z, x z) = a,

and r (p1) = p1, r (p2) = p2, and r (m2) =m (by (1.13.vi)) are distinct prime ideals, so this is an irredundant primarydecomposition of a. We have p1 (m) p2, and p1 6⊆ p2 6⊆ p1, so p1 and p2 are isolated and m is embedded.

6. Let X be an infinite compact Hausdorff space, C (X ) the ring of real-valued continuous functions on X (Chapter 1, Exercise26). Is the zero ideal decomposable in this ring?

No. First recall from [1.16.i] that every maximal ideal m of C (X ) is of the form mx = f ∈C (X ) : f (x) = 0 forsome x ∈X.

Next note that every primary ideal is contained in a unique maximal ideal. Indeed suppose a ⊆ mx ∩my withx 6= y ∈X. Since X is Hausdorff, there are disjoint open neighborhoods U 3 x and V 3 y. Since a compact Hausdorffspace is normal, Urysohn’s lemma can be applied. Thus there are an f ∈C (X ) such that f (x) = 1 and f (X\U ) = 0and a g ∈C (X ) such that f (y) = 1 while f (X\V ) = 0. Since U∩V =∅, we see (X\U )∪(X\V ) =X, so f g = 0 ∈ a,while f ∈my\mx and g n ∈mx\my , so neither is in a, and thus a is not primary.

Now let qi ⊆ mxibe a finite collection of primary ideals and their containing maximal ideals. What follows is

due to Hao Guo [GuoEmail]; the earlier version made an unjustified assumption. Since X is infinite, there is a pointx0 not among the xi , meaning particularly that qi is not contained in mx0

. Each qi , then, contains some fi vanishingat xi but not at x0. It follows that the product of the fi does not vanish at x0, so this product is a nonzero element of∏

qi ⊆⋂

qi . As the finitely many primary ideals qi were arbitrary, there can be no primary decomposition of (0)in C (X ).

63

Ex. 4.7 Chapter 4: Primary Decomposition

7. Let A be a ring and let A[x] denote the ring of polynomials in one indeterminate over A. For each ideal a of A, let a[x]denote the set of all polynomials in A[x] with coefficients in a.i) a[x] is the extension of a to A[x].

Identify a Ã A with its image in A[x]; then ae := aA[x] = a[x]. Very explicitly, a ·A[x] = ∑

ai · pi (x) : ai ∈a, pi (x) ∈A[x], but each ai · pi (x) has all coefficients in a, and so is in a[x]; conversely, each element

∑

ai x i ∈ a[x]can be written as

∑

ai · pi (x) for ai ∈ a and pi (x) = x i ∈A[x].

ii) If p is a prime ideal in A, then p[x] is a prime ideal in A[x].This is [2.7]. To reiterate, note that p[x] is the kernel of the canonical surjection of A[x] onto (A/p)[x], an

integral domain.

iii) If q is a p-primary ideal in A, then q[x] is a p[x]-primary ideal in A[x].q[x] = q+ q · (x), while p[x] = p+ p · (x), and r

(x)

= (x). p and q can be identified with their images in A[x],and in the bigger ring we still have r (q) = p= r (p). Then (1.13.v,iii) give

r

q[x]

= r

r (q)+ r

q · (x)

= r

p+ r (q)∩ r

(x)

= r

p+ p∩ (x)

= r

p+ p · (x)

= r

p[x]

= p[x].

Now to see q[x] is primary, note that the quotient A[x]/q[x]∼= (A/q)[x]. Every zero-divisor in A/q is nilpotent (p.50). Suppose

∑

bi x i = p(x) ∈ (A/q)[x] is a zero-divisor. Then by [1.2.iii], there is a ∈A/q such that a p(x) = 0. Thismeans a · bi = 0 for each i , so each bi ∈ A/q is a zero-divisor, hence nilpotent. Then by [1.2.ii], p(x) is nilpotent.This shows (p. 50) that q[x] is primary.

iv) If a=⋂n

i=1 qi is a minimal primary decomposition in A, then a[x] =⋂n

i=1 qi [x] is a minimal primary decompositionin A[x].

∑

b j x j ∈ a[x] ⇐⇒ ∀ j

b j ∈ a=⋂

i

qi

⇐⇒ ∀i ∀ j (b j ∈ qi ) ⇐⇒ ∀i∑

b j x j ∈ qi [x]

⇐⇒∑

b j x j ∈⋂

i

qi [x],

so a[x] =⋂n

i=1 qi [x] as hoped. By iii), each qi is primary, so this is a primary decomposition. If we leave out someq j , then the above shows

⋂

i 6= j qi [x] =⋂

i 6= j qi

[x], which by assumption is not a[x], using irredundancy of thegiven primary decomposition of a.

v) If p is a minimal prime ideal of a, then p[x] is a minimal prime ideal of a[x].From ii), p[x] is a prime ideal containing a[x]. If q ∈ Spec(A[x]) is such that a[x] ⊆ q ⊆ p[x], then taking

contractions we have a⊆ A∩ q⊆ p, with A∩ q prime. Since p was minimal over a, we have A∩ q= p. Now p⊆ q,so p[x] = p+ p · (x)⊆ q, and thus q= p[x]. Thus p[x] was minimal over a[x].

8. Let k be a field. Show that in the polynomial ring k[x1, . . . , xn] the ideals pi = (x1, . . . , xi ) (1 ≤ i ≤ n) are prime andall their powers are primary.

Write A= k[x1, . . . , xn]. Then A/pi∼= k[xi+1, . . . , xn] is an integral domain, so pi is prime. Now consider some

power pmi . Write qm

i for its intersection with the subring Bi =A[x1, . . . , xi ]. Then pmi = qm

i [xi+1, . . . , xn]. By [4.7.iii],if qm

i is primary, then pmi will also be.

So consider the quotient ring C = Bi/qmi . Write this as C = k[y1, . . . , yi ], where the only relations yi = xi are

(commutativity and) that any monomial in them of total degree greater than m is zero. Consider a product pq inC . If both p and q have nonzero constant term, then so does pq 6= 0. Any term divisible by some y j is annihilatedby q =

∏ij=1 y m−1

j , so p ∈ C is zero-divisor if and only if its constant term is zero. If that is the case, then p m = 0since each term will have total degree ≥ m. Thus every zero-divisor in C is nilpotent and qm

i is primary.

9. In a ring A, let D(A) denote the set of prime ideals p which satisfy the following condition: there exists a ∈ A such that pis minimal in the set of prime ideals containing (0 : a). Show that x ∈A is a zero divisor ⇐⇒ x ∈ p for some p ∈D(A).

If a 6= 0 but xa = 0, then x ∈ (0 : a) 6= (1), so there exists a prime p ⊇ (0 : a) 3 x. By an application of Zorn’sLemma, or by [1.8] applied to A/(0 : a) and (1.1), there is a minimal prime p over (0 : a), which then contains x.

Now suppose x ∈ p ∈ D(A), with a ∈ A such that p ⊇ (0 : a) is minimal. Apparently a 6= 0, since otherwise(1) = (0 : 0) is contained in no prime. By (1.1) and p. 9, p = p/(0 : a) is a minimal prime of A = A/(0 : a), and

64


x = x + (0 : a) ∈ p. Write S = A\p. Since p is minimal, by [3.6], S is maximal in the collection Σ of multiplicativesubmonoids of A not containing 0. Let S ′ be the smallest submonoid of A containing S ∪x; explicitly, S ′ = s xn :s ∈ S, n ≥ 0. Since x ∈ p = A\S, by the maximality of S we have S ( S ′ /∈ Σ, so 0 ∈ S ′\S, and there are s ∈ Sand n > 0 with 0 = s xn = x(s xn−1), showing x is a zero-divisor in A. Write y = s xn−1 6= 0; then xy ∈ (0 : a) in A.Now by assumption y 6= 0, so y /∈ (0 : a), and thus ay 6= 0. But since xy ∈ (0 : a) we have 0= xya = x(ay), so x is azero-divisor in A.

Let S be a multiplicatively closed subset of A, and identify Spec(S−1A) with its image in Spec(A) (Chapter 3, Exercise21). Show that

D(S−1A) =D(A)∩ Spec(S−1A).

Write ae = S−1a for the extension of a Ã A along the canonical φS : A → S−1A and X = Spec(A), S−1X =x ∈ Spec(A) : px ∩ S = ∅. If x ∈ X \S−1X, then pe

x = (1), so not a prime and not in D(S−1A). So evidentlyD(S−1A)⊆ S−1X. From now on suppose p ∈ S−1X.

Let x ∈A. Then (x) is finitely generated, so by (3.15) we have

Ann(x)e = S−1 Ann(x) = S−1(0) : (x)

=

S−1(0) : S−1(x)

=Ann

S−1(x)

=Ann(x/1).

Now for any x ∈ A and s ∈ S we have Ann(x/s) = Ann(x/1) since ax/t s = 0 ⇐⇒ ∃u ∈ S (uax = 0 ∈ A) ⇐⇒ax/s = 0.

Now every ideal of S−1A is extended by (3.11.i), so S−1p ∈D(S−1A) ⇐⇒ ∃x ∈A such that S−1p is minimal overAnn(x/1) = S−1 Ann(x) =. Contracting both sides p = pec ⊇ Ann(x)ec ⊇ Ann(x). If there were an intermediateprime q such that a⊆ q⊆ p, then extending, ae ⊆ qe ⊆ pe , so qe = pe , and by (3.11.iv), q= p. Thus p ∈D(A).

On the other hand, suppose p ∈ D(A)∩ S−1X is minimal over a = Ann(x). Then Ann(x/1) = S−1a ⊆ S−1p. IfS−1q is such that S−1a⊆ S−1q⊆ S−1p, then by (3.11.iv), a⊆ q⊆ p, so by assumption q= p and S−1q= S−1p. ThusS−1p is minimal over S−1a, so S−1p ∈D(S−1A).

If the zero ideal has a primary decomposition, show that D(A) is the set of associated prime ideals of 0.Let the primary decomposition be (0) =

⋂

qi . Then (4.5) and (4.7) say that pi = r (qi ) are the ideals r (0 : x) forx ∈A, so the (finite) set of ideals r (0 : x) are those primes associated to (0). But each r (0 : x) is minimal over (0 : x),so in D(A).

10. For any prime ideal p in a ring A, let Sp(0) denote the kernel of the homomorphism A→Ap. Prove thati) Sp(0)⊆ p.

Since (0)⊆ p, (4.12*.i), (4.12*.vi) twice, and (3.13) give Sp(0) = (0)ec ⊆ pec = p.2

ii) r

Sp(0)

= p ⇐⇒ p is a minimal prime ideal of A.Taking r of i), we see r

Sp(0)

⊆ p regardless of minimality.Now write S =A\p. By (4.12*.i) or [3.1], Sp(0) = x ∈A : ∃s ∈ S (s x = 0)=

⋃

s∈S Ann(s). Thus

p⊆ r

Sp(0)

= r⋃

s∈S Ann(s) p. 9=⋃

s∈S r

Ann(s)

⇐⇒ for each x ∈ p there are n > 0 and s ∈ S such that s xn = 0⇐⇒∀x ∈ p, 0 ∈ (smallest submonoid containing S ∪x)⇐⇒ S is maximal among multiplicative submonoids T 6 3 0 of A⇐⇒[3.6]

p is minimal.

iii) If p⊇ p′, then Sp(0)⊆ Sp′(0).Since p⊇ p′ ⇐⇒ Sp ⊆ Sp′ , this follows from (4.12*.v).

iv)⋂

p∈D(A) Sp(0) = 0, where D(A) is defined in Exercise 9.Obviously, 0 is in each Sp(0). On the other hand, if x 6= 0, then (0 : x) 6= (1), and there is a prime p minimal over

(0 : x); by definition, p ∈D(A). Since (0 : x)⊆ p, there is no s ∈A\p such that s x = 0, and so [3.1] says x /∈ Sp(0).

2 Alternately, suppose x ∈ Sp(0); by (4.12*.i), there is s ∈ Sp such that s x = 0 ∈ p. Since p is prime and doesn’t contain s , we must have x ∈ p.

65


11. If p is a minimal prime ideal of a ring A, show that Sp(0) (Exercise 10) is the smallest p-primary ideal.Write q= Sp(0). Using minimality of p and [4.10.ii], r (q) = p. To see q is primary, suppose xy ∈ q. Then there

is n > 0 such that xn yn = (xy)n ∈ p. If x /∈ q, then xn /∈ p for all n, so yn ∈ p = r (q), and there exists m > 0 suchthat y mn = (yn)m ∈ q. Thus q is primary.

Let a be the intersection of the ideals Sp(0) as p runs through the minimal prime ideals of A. Show that a is containedin the nilradical of A.

Let P ⊆ Spec(A) be the set of minimal prime ideals. Since all primes contains a member of P , we have, by (1.8),that N=

⋂

P . Now for each p ∈ P we have Sp(0)⊆ p by [4.10.i], so a :=⋂

p∈P Sp(0)⊆⋂

P =N.

Suppose that the zero ideal is decomposable. Prove that a= 0 if and only if every prime ideal of 0 is isolated.If (0) is decomposable, then (4.6) states each minimal prime ideal is a minimal ideal of (0) (and vice versa). More-

over, there are only finitely many prime ideals of (0). If every prime ideal of (0) is isolated, then all are minimal, andso (0) =

⋂

primes of (0)=⋂

P =N, forcing a= 0.On the other hand, suppose a = 0. Then 0 = a =

⋂

Sp(0) : p ∈ P gives a primary decomposition of (0). Wemay discard all but finitely many terms to obtain an irredundant decomposition. Now if (0) had an embedded primeP containing an isolated prime p, we would have SP(0)⊆ Sp(0) by [4.10.iii], contradicting irredundancy.

12. Let A be a ring, S a multiplicatively closed subset of A. For any ideal a, let S(a) denote the contraction of S−1a in A. Theideal S(a) is called the saturation of a with respect to S. Prove thati) S(a)∩ S(b) = S(a∩ b).

Note that ae∩be = S−1a∩S−1b= S−1(a∩b) = (a∩b)e by (4.12*.i) and (3.11.v). Then recalling (1.18), S(a)∩S(b) =aec ∩ bec = (ae ∩ be )c = (a∩ b)ec = S(a∩ b).34

ii) S

r (a)

= r

S(a)

. By (1.18), contraction commutes with r , and by (3.11.v), extension S−1 does.Therefore by(4.12*.i), S

r (a)

= r (a)ec =

r (ae )c = r (aec ) = r

S(a)

.5

iii) S(a) = (1) ⇐⇒ a meets S.S(a) = (1) ⇐⇒ 1 ∈ S(a) ⇐⇒ ∃s ∈ S (s = s · 1 ∈ a) ⇐⇒ a∩ S 6=∅.

iv) S1

S2(a)

= (S1S2)(a).If x ∈ (S1S2)(a), then there are s1 ∈ S1 and s2 ∈ S2 such that s1 s2x ∈ a. Then s1x ∈ S2(a), and x ∈ S1

S2(a)

. Onthe other hand, if x ∈ S1

S2(a)

, then there is s1 ∈ S1 such that s1x ∈ S2(a), so there is s2 ∈ S2 such that s2 s1x ∈ a, andthen x ∈ (S1S2)(a).

If a has a primary decomposition, prove that the set of ideals S(a) (where S runs through all multiplicatively closed subsetsof A) is finite.

Let the decomposition be a=⋂

ni=1qi . By (4.9), S(a) is determined entirely by which of the pi = r (qi ) it meets.

This yields at most 2n different possibilities for S(a). (Make at most n possible choices of “empty” or “non-empty,”one for each intersection S ∩ pi .)

13. Let A be a ring and p a prime ideal of A. Then nth symbolic power of p is defined to be the ideal (in the notation ofExercise 12)

p(n) = Sp(pn)

where Sp =A\p. Show thati) p(n) is a p-primary ideal;

3 Alternately, suppose that x ∈ S(a∩ b). Then there is s ∈ S such that s x ∈ a∩ b, so s x ∈ a and s x ∈ b, meaning x ∈ S(a) and x ∈ S(b); thusx ∈ S(a)∩ S(b). Now suppose x ∈ S(a)∩ S(b). Then there are s , t ∈ S such that s x ∈ a and t x ∈ b. Then t (s x) ∈ ta⊆ a and s(t x) ∈ sb⊆ b, so(s t )x ∈ a∩ b. Thus x ∈ S(a∩ b).

4 Note that (4.12*.vi) follows: if a⊆ b, then S(a)∩ S(b) = S(a∩ b) = S(a), so S(a)⊆ S(b).5 Alternately, if x ∈ S

r (a)

, then there is s ∈ S such that s x ∈ r (a), so there is n > 0 such that s n xn = (s x)n ∈ a. Then since s n ∈ S we havexn ∈ S(a), so x ∈ r (S(a)). If x ∈ r (S(a)), then there is n > 0 such that xn ∈ S(a), so there is s ∈ S such that s xn ∈ a. Then (s x)n = s n−1(s x) ∈ aas well, so s x ∈ r (a) and x ∈ S

r (a)

.

66


(3.13) shows pe is maximal, so by (4.2) (pe )n is primary. By (3.11.v), (pe )n = (pn)e . Since contraction preservesbeing primary (p. 50) and p(n) = (pn)ec by (4.12*.i), it follows p(n) is primary.6 As for the radical,

r (p(n))(4.12*.i)= r

(pn)ec (1.18)=

r (pn)ec (3.11.v)= r (pn)ec (1.13.vi)

= pec (3.13)= p.

ii) if pn has a primary decomposition, then p(n) is its p-primary component;First, p(n) is the smallest p-primary ideal containing pn . If q⊇ pn is p-primary and x ∈ p(n), then by (4.12*.i) there

is s ∈ Sp such that x s ∈ pn ⊆ q. Since s /∈ r (q) = p, we have s m /∈ q for all m; as q is primary, it follows that x ∈ q.Thus p(n) ⊆ q.

Let⋂

qi = pn be a primary decomposition. Then by (1.13.vi,iii), p = r (pn) = r⋂

qi

=⋂

r (qi ). By (1.11.ii),p = r (qi ) for some q = qi . Thus p definitely is a prime of pn . If we had r (q j ) ( p for some j , we would havep=

⋂

r (ql )⊆ r (q j )( p, a contradiction; thus p is isolated. Since p(n) is the smallest p-primary ideal containing pn ,pn = p(n) ∩ j 6=i qi is also a primary decomposition; but by the uniqueness (4.11) of isolated primary components, itfollows qi = p(n).

iii) If p(m)p(n) has a primary decomposition, then p(m+n) is its p-primary component;First show p(m+n) is the smallest p-primary ideal containing p(m)p(n). Let q ⊇ p(m)p(n) be p-primary, and x ∈

p(m+n). Then there is s ∈ Sp such that x s ∈ pm+n = pmpn ⊆ p(m)p(n) ⊆ q. Since s /∈ r (q) = p and q is primary, itfollows that x ∈ q. Thus p(m+n) ⊆ q.

Now suppose p(m)p(n) =⋂

qi is a primary decomposition. Since r (p(m)) = r (p(n)) = p, it follows from (1.13.iii,vi)that r (p(m)p(n)) = r (p(m))∩r (p(n)) = p∩p= p. As in part ii), p= r (qi ) for some i , so p is a prime of p(m)p(n), and sincethe radical is not a proper subset of p, we know p is isolated. Since p(m+n) is the smallest p-primary ideal containingp(m)p(n), by (4.11) it follows p(m+n) is the uniquely determined p-primary component of p(m)p(n).

iv) p(n) = pn ⇐⇒ p(n) is p-primary.N.B. The book has a misprint here: by part i), p(n) is always p-primary, so the condition should be that pn is

p-primary.Suppose p(n) = pn . Then since p(n) is p-primary by i), pn is p-primary. On the other hand suppose pn is p-primary.

Then pn =⋂

pn gives a trivial primary decomposition, so by the first uniqueness theorem (4.5), p is the only primeof pn . By part ii), p(n) is the p-primary component of pn , so we conclude p(n) = pn .

14. Let a be a decomposable ideal in a ring A and let p be a maximal element of the set of ideals (a : x), where x ∈ A andx /∈ a. Show that p is a prime ideal belonging to a.

First, let q be p′-primary and x /∈ q, so (q : x) is a p′-primary ideal by (4.4.ii). Suppose x is such that (q : x) ismaximal among all ideals of this form. Let y ∈A\(q : x)⊆A\q. Using (1.12.i,iii) we have (q : x)⊆

(q : x) : y

= (q :xy), and this last is p′-primary by (4.4.ii) since we assumed xy /∈ q. As we are assuming (q : x)maximal of this form,(q : xy) = (q : x), so for every z ∈ A, xy z ∈ q =⇒ x z ∈ q. Taking z = yn , xyn+1 ∈ q =⇒ xyn ∈ q, and stringingthese together, xyn ∈ q =⇒ x ∈ q. But we assumed x /∈ q, so that there is no power yn of y such that yn ∈ (q : x),or in other words y /∈ r (q : x) = p′. Summarizing, y ∈ A\(q : x) =⇒ y ∈ A\p′, or p′ ⊆ (q : x). But by assumption(q : x)⊆ p′, so a maximal ideal of the form (q : x), if such exists, is equal to r (q).

Now write a=⋂n

i=1 qi for an irredundant primary decomposition, and set pi = r (qi ).Suppose x ∈A is such that(a : x) is maximal among such ideals. (1.12.iv) gives (a : x) =

⋂

qi : x

=⋂

(qi : x), and (4.4) shows each (qi : x) is(1) or is pi -primary. We can make any set of (qi : x) = (1), i ∈ I , by taking x ∈

⋂

i∈I qi , so if (a : x) is maximal, whilestill (a : x) 6= (1) (⇐⇒ x /∈ a), we have all but one (qi : x) = (1), and (qi : x)maximal among such proper ideals. Thepreceding paragraph shows this happens if and only if (a : x) = (qi : x) = pi , a prime ideal belonging to a.

15. Let a be a decomposable ideal in a ring A, letΣ be an isolated set of prime ideals belonging to a, and let qΣ be the intersectionof the corresponding primary components. Let f be an element of A such that, for each prime ideal p belonging to a, wehave f ∈ p ⇐⇒ p /∈Σ, and let S f be the set of all powers of f . Show that qΣ = S f (a) = (a : f n) for all large n.

Note that the solution itself doesn’t explicitly use that Σ is composed of isolated primes of a. I think the assump-tion is only needed because if one has an isolated prime p contained in an embedded prime P with p /∈Σ and P ∈Σ,it’s not possible to require f ∈ p /∈Σ but f /∈P ∈Σ.

6 Alternately, suppose xy ∈ p(n); we will show x ∈ p(n) or y ∈ r (p(n)). There is s ∈ Sp such that s xy ∈ pn . If x /∈ p(n), then s x /∈ pn , so thehighest possible power of p containing s x is pn−1. Then, using (1.13.vi), we must have y ∈ p= r (pn).

67


(a : f n)⊆ S f (a): By definition, S f (a) = x ∈A : ∃ f n ∈ S f ( fn x ∈ a)=

⋃

n≥0(a : f n).S f (a) = qΣ: Fix a primary decomposition a =

⋂mi=1 qi , with pi = r (qi ), such that Σ = p1, . . . , pp.7 8 Since pi

are prime, we have S f ∩ pi 6= ∅ just if f ∈ pi just if pi /∈ Σ, so S f meets pp+1, . . . , pm but no others. Then (4.9) saysS f (a) =

⋂pi=1 qi = qΣ.

∃n ≥ 0

qΣ ⊆ (a : f n)

: Since by (1.12.iv) we have (a : f n) =⋂

(qi : f n), we are looking for n large enough thatfor each i and each x ∈ qΣ we have f n x ∈ qi . In case pi ∈Σ, we already have f 0x = x ∈ qΣ ⊆ qi . For pi /∈Σ, we haveby the definition of f that f ∈ pi = r (qi ). Thus there is ni ≥ 1 such that f ni

i ∈ qi . Taking n =maxi ni we then havef n x ∈ qi for all i .

16. If A is a ring in which every ideal has a primary decomposition, show that every ring of fractions S−1A has the sameproperty.

Recall from (3.11.i,ii) that every proper ideal of S−1A is an extended ideal S−1a for some aÃ A with S ∩ a=∅.So let a proper ideal S−1aÃ S−1A be given, and let a=

⋂

qi be a primary decomposition of a. By (3.11.v) we thenhave S−1a = S−1

⋂

qi

=⋂

S−1qi . But by (4.8) that the proper primary ideals of S−1A are exactly those ideals ofthe form S−1q with q primary in A such that S ∩ r (q) = ∅, so

⋂

S−1qi , perhaps omitting a few qi that meet S, is aprimary decomposition of S−1a.

17. Let A be a ring with the following property.(L1) For every ideal a 6= (1) in A and every prime ideal p, there exists x /∈ p such that Sp(a) = (a : x), where Sp =A\p.

Then every ideal in A is an intersection of (possibly infinitely many) primary ideals.For future overuse, we state and prove a slight generalization of a result of [4.11]: if a 6= (1) is an ideal of A and

p is a prime ideal minimal over a, then q = Sp(a) is a p-primary ideal. To prove this, note that in the ring A/a, p/ais a minimal prime, and [4.11] gives that q = Sp/a(0) is p/a-primary. We claim that q = q/a, so that q, being thecontraction of a primary ideal, is primary, by a remark on p. 50. Indeed, if s x ∈ a for s ∈ Sp, then s x = 0, wheres ∈ Sp/a = (A/a)\(p/a), the image of A\p. On the other hand, if s x = 0, with s ∈ Sp/a, then (s+a)(x+a)⊆ a, so s x ∈ a.Finally r (q) = p/a, so using the exercise (1.18) on contraction (along A→A/a), r (q) = r (qc ) = (r (q))c = (p/a)c = p,so q is p-primary as claimed.910

Let p0 be minimal over a0 := a, and set q0 := Sp0(a0). By the above paragraph q0 is p0-primary and by (L1), there

is x0 /∈ p0 such that q0 = (a0 : x0).We claim a0 = q0 ∩ [a0 + (x0)]. Indeed, a0 ⊆ q0 and a0 ⊆ a0 + (x0). On the other hand let a0 + b0x0 ∈ a0 + (x0)

be arbitrary. If it is also in q0 = (a0 : x0), then x0(a0 + b0x0) = a0x0 + b0x20 ∈ a0, implying b0x2

0 ∈ a0 ⊆ q0. Sincex2n

0 /∈ p0 ⊇ q0, which is primary, we have b0 ∈ q0 = (a0 : x0), so b0x0 ∈ a0 and hence a0+ b0x0 ∈ a0.Since this is so, there is an ideal a1 ⊇ a0+(x0)maximal with respect to the requirement that q0∩a1 = a0. Suppose

a1 6= (1). Then there is a prime ideal p1 minimal among those containing a1, and q1 := Sp1(a1) is p1-primary. By (L1),

q1 = (a1 : x1) for some x1 /∈ p1. By the same argument as the previous paragraph, replacing every subscript 0 by a 1,we see a1 = q1 ∩ [a1+(x1)]. Then a0 = q0 ∩ a1 = q0 ∩ q1 ∩ [a1+(x1)].

We continue this process of producing qα by transfinite induction. For the “successor” step, suppose we havea0 = [aα+(xα)]∩

⋂

β≤α qβ for some ordinal α and some primary ideals qβ and aα 6⊆ r (qβ) for anyβ<α. If aα 6= (1),then as above there is aα+1 containing aα+ (xα), maximal subject to the constraint that a0 = aα+1 ∩

⋂

β≤α qβ. Take

7 This paragraph adapted from Yimu Yin’s solution: http://pitt.edu/~yimuyin/research/AandM/exercises04.pdf8 Here is a divergent, worse proof of S f (a) ⊆ qΣ that I came up with before looking up others’ solutions. Assume x ∈ S f (a). By [4.12.i],

S f (a) =⋂

S f (qi ), so there are ni ≥ 0 such that f ni x ∈ S f (qi ) for each i . Since qi is primary, we have x ∈ qi or ( f ni )mi ∈ qi for some ni ≥ 0,meaning f ∈ r (qi ) = pi . By assumption f ∈ pi ⇐⇒ pi /∈Σ, so for each qi with pi ∈Σ we have x ∈ qi . Thus x ∈ qΣ.

9 We can arguably simplify the situation of the problem slightly by replacing A with A/a and a with (0). If we find a way to write⋂

p∈Ξ Sp/a(0) = (0) for some set Ξ ⊆ Spec(A), and these Sp/a(0) are primary, then contracting will show, by (1.18), that a =⋂

p∈Ξ Sp(a) is

an intersection of primary ideals upstairs. Also, (L1) survives in A/a since, by (1.18), (0 : xp)c =

(0)c : (xp)c =

a : (xp) + a

= (a : xp), so(0 : xp) = (0 : xp)

c e = (a : xp)e = Sp(a)/a. (Finally looking over the book’s “hint,” I’m no longer sure this simplification really simplifies very

much. In fact, it may make things a little harder.)10 What I now want to do is as follows. Recall from [4.9] the set D(A) of prime ideals p of A such that there exists a ∈ A with p minimal in

the set of prime ideals containing (0 : a). By [4.10.iv],⋂

p∈D(A) Sp(0) = 0. This would be what we wanted if we could be assured the Sp(0) wereall are primary. Let P be the set of minimal primes of A. If p ∈ P , then ([4.11]) Sp(0) = (0 : xp) is p-primary. But it’s not clear we should haveD(A) = Spec(A) or D(A) = P . We also have

⋂

p∈P Sp(0) =⋂

p∈P

0 : (xp)

=

0 :∑

p∈P (xp)

, which will equal zero if∑

p∈P (xp) contains anon-zero-divisor. But I am not sure why this would be the case.

68

http://pitt.edu/~yimuyin/research/AandM/exercises04.pdf


pα+1 a minimal prime over aα+1, so that qα+1 = Spα+1(aα+1) is pα+1-primary. By (L1) there is xα+1 /∈ pα+1 such that

qα+1 = (aα+1 : xα+1). The same argument showing a0 = q0 ∩ [a0+(x0)] shows that aα+1 = qα+1 ∩ [aα+1+(xα+1)], so

a0 = aα+1 ∩⋂

β≤α

qβ = [aα+1+(xα+1)]∩⋂

β≤α+1

qβ.

For the “limit” step, suppose that for all α < β we have a0 = aα+1 ∩⋂

γ≤α qγ , with aα ( aγ for α < γ . Setaβ :=

⋃

α<β aα. Then aα ( aβ for each α <β, and

aβ ∩⋂

α<β

qα =⋃

α<β

aα ∩⋂

α<β

qα

=⋃

α+1<β

aα+1 ∩⋂

γ≤αqγ ∩

⋂

α<γ<β

qγ

=⋃

α+1<β

a0 ∩⋂

γ<β

qγ

=⋃

α+1<β

a0 = a0.

If at any stage we get aα = (1) we are done. In particular, if for some n < ω we have an = (1), then we havefound a primary decomposition of a = a0. Since for α < β we have strict containment aα ( aβ, the number ofdifferent aα we encounter is bounded by the cardinality of A, so the process does eventually terminate, leaving uswith a decomposition a=

⋂

α<β qα for some primary ideals qα and some β of cardinality less than that of A.

18. Consider the following condition on a ring A:(L2) Given an ideal a and a descending chain S1 ⊇ S2 ⊇ · · · ⊇ Sn ⊇ · · · of multiplicatively closed subsets of A, there existsan integer n such that Sn(a) = Sn+1(a) = · · · . Prove that the following are equivalent:i) Every ideal in A has a primary decomposition;ii) A satisfies (L1) and (L2).

i) =⇒ (L1): Let aÃA have primary decomposition⋂

qi and let p ∈ Spec(A) and Sp =A\p. Also let pi = r (qi ).LetΣ be the set of indices i with qi ⊆ pi ⊆ p (hence qi∩Sp =∅) andΞ the set of those with qi 6⊆ p (hence qi∩Sp 6=∅).(4.9) says that Sp(a) =

⋂

i∈Σ qi . Now (a : x) =⋂m

i=1(qi : x), and (qi : x) = (1) if x ∈ qi and = qi if x /∈ pi , so we arelooking for x ∈

⋂mi∈Ξ qi \

⋃

i∈Σ pi . For each i ∈ Ξ, there is an element xi ∈ qi \p, and then x :=∏

i∈Ξ xi ∈⋂

qi \psince p is prime. But since

⋃

i∈Σ pi ⊆ p, we have (a : x) = Sp(a) as desired.i) =⇒ (L2): Let a Ã A have primary decomposition

⋂mi=1 qi . (4.9) states that Sn(a) =

⋂

i∈Ξnqi , where Ξn ⊆

1, . . . , m is the set of indices i such that Sn ∩ r (qi ) =∅. As n gets bigger, Ξn decreases; but it can decrease at mostm times, so at some finite stage it is all done decreasing and Sn(a) has stabilized.

ii) =⇒ i): Let aÃA be given. [4.17] shows there exist primary qα, α <β, such that a=⋂

qα. We will be done ifwe can show finitely many suffice. Write pα = r (qα) and Sβ =A\

⋃

α<β pα. Then [3.7.i] shows that the Sα are saturatedmultiplicative submonoids of A and apparently Sα ⊇ Sβ for α <β. Recall from the proof of [4.17] that at each finitestage of the construction we have an ideal an+1 such that a = an+1 ∩ qn ∩ · · · ∩ q0, and an+1 6⊆ pm for m < n + 1.Then an+1 ∩ Sn 6=∅ and [4.12.iii] says that Sn(an+1) = (1), so Sn(a) = q0 ∩ · · · ∩qn . (L2) says that the sequence Sn(a)stabilizes in finitely many steps, say at Sn(a) = q0 ∩ · · · ∩ qn . Then for all α≥ n we have Sα(a) = q0 ∩ · · · ∩ qn .

Write a = Sn(a) ∩ qα ∩⋂

γ /∈0, ..., n,α qγ . Then taking Sα and using the finitary distributive property [4.12.i],we should get a term Sα(qα). Note qα ⊆ pα does not meet A\pα ⊇ Sα. Thus if x ∈ Sα(qα), so there is s ∈ Sαsuch that s x ∈ qα, that qα is primary implies ∃n (s n ∈ qα) (which cannot happen) or x ∈ qα, so the contributedterm is Sα(qα) = qα. Thus

⋂ni=0 qi ∩ qα ∩ · · · =

⋂ni=0 qi , so

⋂ni=0 qi ⊆ qα. Since this holds for all α > n, we see

a=⋂

α<β qα =⋂n

i=0 qi has a primary decomposition.

19. Let A be a ring and p a prime ideal of A. Show that every p-primary ideal contains Sp(0), the kernel of the canonicalhomomorphism A→Ap.

This was proved for the first statement of [4.11].

Suppose that A satisfies the following condition: for every prime ideal p, the intersection of all p-primary ideals of Ais equal to Sp(0). (Noetherian rings satisfy this condition: see Chapter 10.) Let p1, . . . , pn be distinct prime ideals, none ofwhich is a minimal prime ideal of A. Then there exists an ideal a in A whose associated prime ideals are p1, . . . , pn .

We attempt an induction proof. For n = 1, the ideal a = p1 satisfies the condition. Suppose that the result hasbeen proved for n, and let pn+1 be one last prime. Then there is an ideal b with a primary decomposition

⋂ni=1 qi

where r (qi ) = pi . What we’d like is to take any pn+1-primary ideal qn+1 and let a = b∩ qn+1. This will give us thedecomposition we want unless it is redundant. If it is redundant, there is some term containing the intersection of

69


the rest. Taking radicals and using the distributivity property (1.13.iii), the intersection of some n of the primes iscontained in the remaining prime. By (1.11.ii), this implies that the big prime contains one of the first n. We canarrange then, by reordering the pi so that pn+1 is maximal among them, that qn+1 we don’t have

⋂

i 6= j qi ⊆ q j forany j 6= n+ 1. Write p′ = pn+1

Now we only have to worry about the possibility that⋂n

i=1 qi ⊆ q′. If there is a p′-primary q′ for which thisdoesn’t happen, we are done. Otherwise, taking the intersection of both sides over all such, we see

⋂ni=1 qi ⊆ Sp′(0),

using the first part of [4.11]. By [4.10.iii], for a smaller prime p⊆ p′ we have⋂n

i=1 qi ⊆ Sp′(0)⊆ Sp(0); in particular,we may take p minimal. Now again taking radicals of both sides and using (1.13.iii), we get

⋂ni=1 pi ⊆ p. By (1.11.ii)

again, this means pi ⊆ p for some i ∈ 1, . . . , n; but then by minimality of p, we have pi = p. But the assumption ofthe problem was that none of the pi were minimal, so the possibility that worried us in this paragraph is forestalled,and we are done.

Primary decomposition of modulesPractically the whole of this chapter can be transposed to the context of modules over a ring A. The following exercises

indicate how this is done.

20. Let M be a fixed A-module, N a submodule of M . The radical of N in M is defined to be

rM (N ) = x ∈A : xq M ⊆N for some q > 0.Show that rM (N ) = r (N : M ) = r

Ann(M/N )

. In particular, rM (N ) is an ideal.

x ∈ rM (N ) ⇐⇒ ∃q > 0 (xq M ⊆N ) ⇐⇒ ∃q > 0

xq ∈ (N : M )

⇐⇒ x ∈ r (N : M ).

By (2.2.ii), and since N ⊆ M , we have (N : M ) = Ann

(N +M )/N

= Ann(M/N ), so taking radicals, r (N : M ) =r

Ann(M/N )

.

State and prove the formulas for rM analogous to (1.13).−i) P ⊆N =⇒ rM (P )⊆ rM (N ): xq M ⊆ P , implies xq M ⊆N .0) rB (C

n) = rB (C ) for B an A-algebra, C a subalgebra, and n > 0: C n ⊆C , so rB (Cn)⊆ rB (C ) by −i).

If xq B ⊆C , then taking nth powers,and remembering 1 ∈ B , gives xqnB ⊆C n .

i) rB (b)⊇ f −1(b), for f : A→ B an A-algebra and bÃ B: If b ∈ b, then bB = (b )⊆ b,and by definition if f (a) = b , then aB = f (a)B ⊆ b.

ii) r

rM (N )

= rM (N ): x ∈ r

rM (N

) ⇐⇒ ∃p > 0

x p ∈ rM (N )

⇐⇒ ∃p, q > 0

x pq M ⊆N

⇐⇒ x ∈ rM (N ).iii) rM (N ∩ P ) = rM (N )∩ rM (P ): If xn M ⊆N ∩ P , then trivially xn M ⊆N and xn M ⊆ P .

If xn M ⊆N and x p M ⊆ P , then for q =maxn, p we have xq M ⊆N ∩ P .iv) rM (N ) = (1) ⇐⇒ M =N : 1 ∈ r

Ann(M/N )

⇐⇒ 1 ∈Ann(M/N ) ⇐⇒ M/N = 0 ⇐⇒ M =N .v) rM (N + P )⊇ r

rM (N )+ rM (P )

: By −i), rM (N ), rM (P )⊆ rM (N + P ), so rM (N )+ rM (P )⊆ rM (N + P ).Taking radicals and applying ii), r

rM (N )+ rM (P )

⊆ r

rM (N + P )

= rM (N + P ).The converse is false. Let A 6= 0, M =A⊕A with action a(b , c) = (ab , ac),

N =A⊕ (0), and P = (0)⊕A,Then M =N + P , so rM (N + P ) = (1), but rM (N ) = rM (P ) = 0.

vi) If p is prime, r (pn) = p for all n > 0: I’ve no clue how to interpret a power of a module.I seem to have failed this problem, in that I wasn’t sure in all cases what the appropriately analogous formulas

were. The ones involving algebras were stretches, brought about by the difficulty of comparing N and rM (N ), onebeing a submodule of M and the other being an ideal of A.

21. An element x ∈A defines an endomorphism φx of M , namely m 7→ x m. The element x is said to be a zero-divisor (resp.nilpotent) in M if φx is not injective (resp. is nilpotent). A submodule Q of M is primary in M if Q 6= M and everyzero-divisor in M/Q is nilpotent.

Show that if Q is primary in M , then (Q : M ) is a primary ideal and hence rM (Q) is prime ideal p. We say that Q isp-primary (in M ).

Suppose xy ∈ (Q : M ) and y /∈ (Q : M ) so that xy(M/Q) = 0 but y(M/Q) 6= 0. Then the endomorphismφx φy = φxy of M/Q is zero, though φy is not. Then φx has non-empty kernel, so x is a zero-divisor of Q. By

70


the definition of “primary,” x is then nilpotent in M/Q, so for some n > 0 we have 0= φnx = φxn : M/Q→M/Q.Then the associated endomorphismφxn : m 7→ xn m of M has image in Q, so xn ∈ (Q : M ). Thus (Q : M ) is primary.

Prove the analogues of (4.3) and (4.4).Lemma 4.3*. If Qi ⊆M (1≤ i ≤ n) are p-primary, then Q =

⋂ni=1 Qi is p-primary.

Since the Qi are primary, hence not equal to M , their intersection Q 6= M . Suppose x ∈ A is a zero-divisor ofM/Q. Then there is some nonzero m ∈M such that x m ∈Q =

⋂

Qi . Then x is a zero-divisor of each M/Qi , so byassumption is nilpotent, meaning there is ni > 0 such that xni M ⊆Qi . Taking n =maxi ni we see xn M ⊆Q, so x isnilpotent in M/Q. Thus Q is primary. As for the radical,

r (Q : M ) = r

⋂

Qi : M

(1.12.iv)= r

⋂

(Qi : M )

(1.13.iii)=

⋂

r (Qi : M ) =⋂

i

p= p.

Lemma 4.4*. i) Let N ⊆M be A-modules and m ∈N. Then (N : m) = (1);ii) Let Q ⊆M be a p-primary submodule, and m ∈M . If m /∈Q then (Q : m) is p-primary;iii) Let Q ⊆M be a p-primary submodule, and x ∈A. If x /∈ p then (Q : x) := m ∈M : x m ∈Q=Q.

i): Since m ∈N and N is an A-module, Am ⊆N , so (N : m) = (1).ii): Suppose xy ∈ (Q : m), so xy m ∈ Q. Suppose y /∈ (Q : m), so that y m /∈ Q. Then x is a zero-divisor in

M/Q, so by the assumption Q is primary, there is n > 0 such that xn acts as zero on M/Q. Then xn M ⊆Q, and inparticular xn m ∈Q, so xn ∈ (Q : m). Thus (Q : m) is primary.

Note that m ∈ M implies (Q : M ) ⊆ (Q : m). Taking radicals, p ⊆ r (Q : m). On the other hand assumex ∈ r (Q : m). Then for some minimal n > 0 we have xn m ∈Q. Then x(xn−1m) = 0 in M/Q, so x is a zero-divisorof M/Q and hence there is p > 0 such that x p M ⊆Q. Then x ∈ rM (Q) = p. Thus (Q : m) is p-primary.

iii): Obviously if m ∈Q then x m ∈Q, so Q ⊆ (Q : x). By contraposition, we will suppose m ∈ (Q : x)\Q andshow x ∈ p. Well, x m ∈Q, and m 6= 0 in M/Q, so x is a zero-divisor, and for some power n > 0 we have xn M ⊆Q.But then x ∈ rM (Q) = p.

22. A primary decomposition of N in M is a representation of N as an intersection

N =Q1 ∩ · · · ∩Qn

of primary submodules of M ; it is a minimal primary decomposition if the ideals pi = rM (Qi ) are all distinct and ifnone of the components Qi can be omitted from the intersection; that is Qi 6⊇

⋂

j 6=i Q j (1≤ i ≤ n).Prove the analogue of (4.5), that the prime ideals pi depend only on N (and M ). They are called the prime ideals

belonging to N in M .

Theorem 4.5*. Let N be a decomposable submodule of M and let N =⋂n

i=1 Qi be a minimal primary decompositionof N . Let pi = rM (Qi ) (1 ≤ i ≤ n). Then the pi are precisely the prime ideals which occur in the set of ideals r (N : m)(m ∈M ), and hence are independent of the particular decomposition of N .

Set Pi =⋂

j 6=i Q j . By the assumption of irredundancy, Qi ( Pi . Let m ∈ Pi\Qi , and consider the ideal (N : m) =

(Pi ∩Qi : m)(1.12.iv)= (Pi : m)∩ (Qi : m). By (4.4*.i,ii) of [4.21] above (Pi : m) = M and (Qi : m) is pi -primary, so

(N : m) is pi -primary. Thus each pi is r (N : m) for some m ∈M .11

Suppose on the other hand that r (N : m) is a prime p for some m ∈ M . Note (N : m) =

⋂

Qi : m

(1.12.iv)=

⋂

(Qi : m), so by (4.4*) above, p = r (N : m) =⋂

m /∈Qipi . Since the prime p is an intersection of some of the pi ,

(1.11.ii) shows p= pi for some i .

Show that they are also the prime ideals belonging to 0 in M/N.Note that for any module P ⊆M we have (N : P ) = (0+N : P+N ). Indeed xP ⊆N ⇐⇒ x(P+N )⊆ 0+N =N .

Taking radicals, r (N : P ) = r (0+N : P +N ). Specializing to cyclic submodules Am gives r (N : m) = r (0 : m), soone is prime just if the other is, and by Theorem 4.5*, the same primes belong to N ⊆M and 0⊆M/N .

11 I owe this part of the argument to Multiplicative Theory of Ideals by Max D. Larsen and Paul Joseph McCarthy. I had initially started reasoningabout ideals of the form

(Q : M ) : x

, and was trying to prove that if N =⋂

Qi is an irredundant decomposition, then (N : M ) =⋂

(Qi : M )waslikewise.

71


23. State and prove the analogues of (4.6)–(4.11) inclusive. (There is no loss of generality in taking N = 0.)We must convince ourselves we genuinely aren’t losing any generality. What we should do is try to lift a primary

decomposition, as we know (p. 50) primary ideals are preserved under contraction. So suppose we are given an irre-dundant primary decomposition 0=

⋂

Qi/N in M/N (recalling the correspondence (p. 18) between submodules ofM/N and submodules of M containing N ). Apparently N =

⋂

Qi , and we should show that the Qi are rM/N (Qi/N )-primary. Much as in the last part of [4.22], we have (Qi/N : M/N ) = x ∈ A : xM ⊆ Qi = (Qi : M ), and takingradicals gives rM (Qi ) = rM/N (Qi/N ). Now we must show Qi is primary. Suppose x ∈ A is a zero-divisor of M/Qi .The third isomorphism theorem (2.1.i) gives M/Qi

∼= (M/N )/(Qi/N ), so x is a zero-divisor of the latter, hencenilpotent since Qi/N is primary, hence nilpotent in M/Qi since they are isomorphic. Thus Qi is primary. It is clearthat irredundancy is preserved under lifting, as Qi/N ⊇

⋂

j 6=i Q j/N ⇐⇒ Qi ⊇⋂

j 6=i Q j by the order-preservingcorrespondence of p. 18.12

Proposition 4.6*. Let N ⊆ M be a decomposable module. Then any prime ideal p⊇ rM (N ) contains a minimal primeideal belonging to N, and thus the minimal prime ideals of N are precisely the minimal elements in the set of all primeideals containing rM (N ).

Write N =⋂

Qi , so that ([4.20.iii]) rM (N ) =⋂

rM (Qi ) =⋂

pi . If p ⊇⋂

pi , then by (1.11.ii), there is i withp ⊇ pi , and surely for any p j ⊆ pi we have p j ⊆ p, so p contains an isolated prime ideal of N . In particular, if p isminimal over rM (N ), this shows it equals some isolated p j .

Proposition 4.7*. Let N ⊆ M be a decomposable module, let N =⋂n

i=1 Qi be a minimal primary decomposition, andlet pi = rM (Qi ). Then

n⋃

i=1

pi = x ∈A : (N : x) 6=N.

In particular, if 0⊆ M is decomposable, the set D ⊆ A of zero-divisors of M is the union of the prime ideals belonging to0.

Since by [4.22], the set of primes associated to N ⊆ M is the same as that associated to 0⊆ M/N , and for x ∈ Awe have (N : x) 6=N ⊆M just if (0 : x) 6= 0⊆M/N , we can indeed assume N = 0.

Then the right-hand side D is the set of x such that there exists m 6= 0 ∈M such that m ∈ (0 : x), or x m = 0; withis to say D is the set of zero-divisors of M . Now if x ∈ r (D), then there is a nonzero m ∈ M and a least n > 0 such

that xn m = 0. Then m′ = xn−1m 6= 0 and x m = 0, so x ∈D . Thus D = r (D) = r

⋃

m 6=0(0 : m)

=⋃

m 6=0 r (0 : m).

The proof of Theorem 4.5* ([4.22]) shows that each r (0 : m) for m 6= 0 is the intersection of some of the pi , andeach pi = r (0 : m) for some m. Thus D =

⋃

pi .

Proposition 4.8*. Let S be a multiplicative submonoid of A, and let Q ⊆M be a p-primary module.i) If S ∩ p 6=∅, then S−1Q = S−1M .ii) If S ∩ p=∅, then S−1Q is a S−1p-primary submodule of S−1M , and its preimage ( contraction) under the canonicalmap M → S−1M is Q. Hence primary S−1A-submodules of S−1M correspond to primary A-submodules of M .

i): Let s ∈ S ∩ p. Since p = rM (Q), there is n > 0 such that s n M ⊆ Q. Then any element m/t ∈ S−1M can bewritten as s n m/s n t ∈ S−1Q.

ii): Suppose x/s ∈ S−1A is a zero-divisor in S−1M/S−1Q. Then there is some non-zero m/t ∈ S−1M/S−1Q suchthat (x/s)m/t = xm/st = 0. Then xm/st ∈ S−1Q, so there is u ∈ S such that u x m ∈ Q ⊆ M . Then since m/twas non-zero, m /∈Q, so u x is a zero-divisor of M/Q, hence nilpotent since Q is primary. Then there is n > 0 suchthat (u x)n M ⊆Q. That means xn S−1M = xn un S−1M ⊆ S−1Q, so x is nilpotent in S−1M/S−1Q, meaning S−1Q isprimary.

If x ∈ p = rM (Q), let n > 0 be such that xn M ⊆ Q. Then for arbitrary s ∈ S we have (x/s)n S−1M ⊆ S−1Q, sox/s ∈ r (S−1Q : S−1M ) = rS−1M (S

−1Q). On the other hand, if x/s ∈ rS−1M (S−1Q) then (xn/1)S−1M = (x/s)n S−1M ⊆

S−1Q for some n > 0. Thus for every m/t ∈ S−1M we have xn m/s t ∈ S−1Q. This means there is u ∈ S such thatu xn m ∈ Q. Then u xn is a zero-divisor of M/Q, so nilpotent in M/Q, and so some power takes M into Q, andu xn ∈ rM (Q) = p. But then since p is prime and u /∈ p we have xn ∈ p, so x ∈ r (p) = p, and finally x/s ∈ S−1p.Therefore S−1Q is S−1p-primary.

12 Note that, on the other hand primary decomposition does not generally survive the quotient process. Indeed, Example 3) on p. 51 showsthat the primary ([4.8]) ideal (x, z)2 of k[x, y, z], where k is a field, has image no longer primary in the quotient ring k[x, y, z]/(xy − z2).

72


Now suppose m ∈M is such that m/1 ∈ S−1Q. Then there is s ∈ S such that s m ∈Q. By (4.7*) above, p= x ∈A : Q 6= (Q : x), so m ∈ (Q : s) =Q as s /∈ p.

Finally, we show that every S−1A-submodule N ′ of S−1M is an extended module of the form S−1N for someA-submodule N ⊆ M . Indeed, let N be the set of n ∈ M such that n/1 ∈ N ′, (which we can also think of as thecontraction of N along the canonical map f : M → S−1M ). If n/s ∈N ′, then n/1= s(n/s) ∈N ′, so n ∈N and hencen/s ∈ S−1N . On the other hand f (N ) = f

f −1(N ′)

⊆N ′, so S−1N ⊆N ′.

Proposition 4.9*. Let S be a multiplicative submonoid of A and let N ⊆M be a decomposable ideal. Let N =⋂n

i=1 Qibe a minimal primary decomposition of N . Let pi = rM (Qi ) and suppose the Qi numbered so that S meets pp+1, . . . , pn

but not p1, . . . , pp . Write S(N ) = m ∈M : m/1 ∈ S−1N. Then

S−1N =p⋂

i=1

S−1Qi , S(N ) =p⋂

i=1

Qi ,

and these are minimal primary decompositions.By (3.4.ii) we have S−1N =

⋂ni=1 S−1Qi . By (4.8*.i) above, we have S−1Qi = S−1M for i > p, so S−1N =

⋂pi=1 S−1Qi , and by (4.8*.ii), S−1Qi is S−1pi -primary for i ≤ p. Since these pi don’t meet S, by (3.11.iv), the

S−1pi are distinct primes of S−1A. If we had, for some j ≤ p, that S−1Q j ⊇⋂p

j 6=i=1 S−1Qi , then taking preim-

ages under f : M → S−1M we see that Q j ⊇⋂

j 6=i Qi , contradicting the assumed irredundancy of the Qi . Thus

S−1N =⋂p

i=1 S−1Qi is an irredundant primary decomposition of S−1N . Taking preimages under f : M → S−1M ,

S(N ) = f −1(S−1N ) = f −1

p⋂

i=1

S−1Qi

=p⋂

i=1

f −1(S−1Qi ) =p⋂

i=1

Qi

by (4.8*.ii) again. This is an irredundant primary decomposition since the decomposition of N is.

Theorem 4.10*. Let N ⊆ M be a decomposable ideal, let N =⋂n

i=1 Qi be a minimal primary decomposition of N , letpi = rM (Qi ), and let Σ = pi1

, . . . , pim be an isolated set of prime ideals of N . Then

⋂

pi∈ΣQi is independent of the

decomposition.Let S =A\

⋃

Σ. Then S is a multiplicative submonoid, and for p ∈Σwe have p∩S =∅, while if p /∈Σ, then sincep is not contained in an element of Σ by isolation, (1.11.i) shows p 6⊆

⋃

Σ, so p∩ S 6=∅. Then⋂

pi∈ΣQi = S(N ) by

(4.9*), so this intersection is actually independent of the Qi chosen.

Corollary 4.11*. The isolated primary components (i.e., the primary components Qi corresponding to minimal primeideals pi ) are uniquely determined by N.

Let pi be an isolated prime of N . TakingΣ= pi and Spi=A\pi in (4.10*) above gives Spi

(N ) =Qi independentof the choice of decomposition.

73

Chapter 5: Integral Dependence and Valuations

EXERCISES1. Let f : A→ B be an integral homomorphism of rings. Show that f ∗ : Spec(B)→ Spec(A) is a closed mapping, i.e. that it

maps closed sets to closed sets. (This is a geometrical equivalent of (5.10).)

Write C = f (A). That f is integral means that B is integral over C . Write f : Ap− C

i,→ B . f will be closed if

both p and i are closed.But p∗ is a homeomorphism between Spec(C ) and V

ker(p)

⊆ Spec(A) by [1.21.v]. For any closed subsetK ⊆ Spec(C ), we have p∗(K) closed in the subspace topology on V

ker(p)

. Since this subset of Spec(A) is closed aswell, p∗(K) is closed in Spec(A). Thus all surjections induce closed maps on prime spectra.

For closedness of i∗, any closed subset of Spec(B) is ([1.15.i]) the set V (b) of all primes containing some radicalideal bÃ B . Let c= i∗(b) = b∩C , we claim i∗

V (b)

=V (c). If b⊆ q ∈ Spec(B), then intersecting both sides withC gives c ⊆ i∗(q) ∈ Spec(C ), so i∗

V (b)

⊆ V (c). Surjectivity is a little less obvious. Every prime p ⊇ c induces aquotient prime p of C/c. (5.6.i) says j : C/c B/b is integral, so by (5.10), there is there is q ∈ Spec(B/b) (the imageof q ∈ Spec(B)) such that j ∗(q) = p. Thus j ∗ surjectively maps Spec(B/b) ≈ V (b) to Spec(C/c) ≈ V (c), so using[3.21.iii], i∗

V (b)

=V (c) and i∗ is closed.1

2. Let A be a subring of a ring B such that B is integral over A, and let f : A → Ω be a homomorphism of A into analgebraically closed field Ω. Show that f can be extended to a homomorphism of B into Ω.Ω is an integral domain, so f (A) is as well. Thus p= ker( f ) is a prime ideal of A, and f (A)∼= A/p. By Theorem

5.10, there is a prime ideal qÃ B with q∩A= p. Then B/q, by (5.6.i), is integral over A/p. Thus it suffices to provethe result in the case A⊆ B are integral domains with B integral over A and f : AΩ is an injection.

We use Zorn’s Lemma to construct an embedding B Ω. Let Σ be the set of pairs (C , σ), where A⊆ C ⊆ B/and σ : C → Ω is an embedding, and such that σ |A = f . Partially order Σ by (C , σ) ≤ (C ′, σ ′) just if C ⊆ C ′ andσ = σ ′|C . Σ 6=∅ since the inclusion (A, f ) is a minimal element. If

(Cα, σα)

αis a chain in Σ, then

⋃

Cα ⊆ C andσ =

⋃

σα is a well defined homomorphism, injective since each σα is, so every chain has an upper bound. By Zorn’sLemma, there is a maximal element (C , σ) ∈Σ. We will be done if C = B .

Suppose for a contradiction then there is b ∈ B\C , and b is integral over A, hence a fortiori over C . Say b satisfiesp(x) =

∑

ci x i ∈C [x], and write (σp)(x) =∑

σ(ci )xi ∈Ω[x] Then the expected composition

C [x]→Ω[x]→Ω[x]/

(σp)(x)∼=Ω

has kernel

p(x)

, and so descends to an injection C [x]/

p(x)

Ω restricting to σ on C . But C [x]/

p(x)∼=C [b ],

and by assumption C (C [b ]⊆ B so the induced map C [b ]Ω contradicts maximality of σ .

3. Let f : B→ B ′ be a homomorphism of A-algebras, and let C be an A-algebra. If f is integral, prove that f ⊗1: B⊗AC →B ′⊗AC is integral. (This includes (5.6) ii) as a special case.)

Start with a decomposable element x⊗c ∈ B ′⊗AC . Since x is integral over B , it satisfies some polynomial equation∑n

i=0 f (bi )xi = 0 for bi ∈ B (with leading coefficient bn = 1). Then

∑

f (bi )xi ⊗ c n = 0 in B ′ ⊗AC . Rearranging,

0=∑

f (bi )⊗ c n−i

(x⊗ c)i . But each f (bi )⊗ c n−i ∈ im( f ⊗ idC ), so x⊗ c is integral over im( f ⊗ idC ). But elementof B ′⊗AC is a finite sum of elements of the form x ⊗ c , so (5.3) shows that the whole ring B ′⊗AC is integral overim( f ⊗ idC ).

The parenthetical comment follows from setting C = S−1A and using (3.5), which states S−1A⊗AB ∼= S−1B .

1 I went this quotient route because first applying (5.10) for p ∈V (c) gives a prime q ∈ Spec(B), but it wasn’t completely obvious to me thatq ∈V (b).

74

Chapter 5: Integral Dependence and Valuations Ex. 5.4

4. Let A be a subring of B such that B is integral over A. Let n be a maximal ideal of B and let m= n∩A be the correspondingmaximal ideal of A (see (5.8)). Is Bn necessarily integral over Am?

Per the book’s suggestion, no. Let k be a field, of characteristic 6= 2 so that x + 1 6= x − 1, B = k[x], andA = k[x2 − 1]. Then x ∈ B satisfies y2 − [1 − (x2 − 1)] = 0 in A[y], so is integral over A. Then by (5.3), B isintegral over A. Now consider the ideal n := (x − 1) of B , and let m := n ∩ A = (x2 − 1). Then Sm := A\m. Asx + 1 /∈ n we have 1/(x + 1) ∈ Bn. If 1/(x + 1) were integral over Am, we would have ai/si ∈ S−1

m A (in least terms)such that

∑ni=0[ai/si ][1/(x+1)]i = 0 in Bn and an/sn = 1. We can rewrite that as

∑

ai (x+1)n−i/si (x+1)n = 0, and,B being an integral domain, we can multiply through by (x + 1)n

∏ni=0 si to get

∑

ai ti (x + 1)n−i = 0 in B , whereti =

∏

j 6=i s j ∈A\(x2−1) and an = 1. Then (x+1) divides each term but possibly an tn(x+1)0 = tn . Since 0 ∈ (x+1),this forces tn ∈ (x + 1)∩A= (x2− 1), a contradiction.

5. Let A⊆ B be rings, B integral over A.i) If x ∈A is a unit in B then it is a unit of A.

Let x ∈ A∩ B×. Then x is not in any n ∈ Max(B). Let m ∈ Max(A). By (4.10), there is n ∈ Spec(B) such thatn∩A=m, and by (5.8), n is maximal. Thus x /∈ n, hence x /∈m. As m was arbitrary, x is in no maximal ideal of A,and hence x ∈A×.2

ii) The Jacobson radical of A is the contraction of the Jacobson radical of B.For a maximal ideal m of A, (5.10) and (5.8) give a unique maximal ideal n of B such that n ∩ A = m. Write

N ⊆Max(B) for the set of these. Then

R(A) =⋂

Max(A) =⋂

n∈N

(A∩n) =A∩⋂

N ⊇A∩⋂

Max(B) =A∩R(B).

On the other hand, by (5.8) again, the set M = n∩A : n ∈Max(B) is a subset of Max(A). Thus

A∩R(B) =A∩⋂

Max(B) =⋂

n∈Max(B)

(A∩n) =⋂

M ⊇R(A).

6. Let B1, . . . , Bn be integral A-algebras. Show that∏n

i=1 Bi is an integral A-algebra.Use induction, and assume we have the n = 2 case. The base step n = 1 is trivial, so suppose we have proved the

proposition up to n and have B1, . . . , Bn+1 integral A-algebras. The inductive assumption yields that∏n

i=1 Bi is anintegral A-algebra, and the n = 2 case shows

∏n+1i=1 Bi

∼=∏n

i=1 Bi ×Bn+1 is an integral A-algebra as well.So it is now enough to prove the n = 2 case. Let B and C be integral A-algebras, and (b , c) ∈ B×C . Then3, 4there

are p(x) ∈A[x] such that p(b ) = 0 in B and q(x) ∈A[x] such that q(c) = 0 in C . Therefore p

(b , c)

=

0, p(c)

andq

(b , c)

=

q(b ), 0

in B ×C , so multiplying these, (pq)

(b , c)

= p

(b , c)

q

(b , c)

=

0, p(c)

q(b ), 0

= (0, 0),where (pq)(x) has leading coefficient 1, showing (b , c) is integral over the image of A.

7. Let A be a subring of a ring B, such that the set B\A is closed under multiplication. Show that A is integrally closed in B.Let b ∈ B be integral over A, and let n ≥ 2 be such that b satisfies an equation b n+an−1b n−1+ · · ·+a1b +a0 = 0

for ai ∈A. Since 0, a0 ∈A, we have b n + · · ·+ a1b ∈A. We can factor this as b (b n−1+ · · ·+ a1) ∈A, and since B\A ismultiplicatively closed, either b ∈ A or b n−1+ an−1b n−2+ · · ·+ a1 ∈ A. Iterating this process, we eventually arriveat b + an−1 ∈A, so b ∈A.

2 The proof from (5.7) also works without modification. Suppose x−1 ∈ B . By uniqueness of inverses in B , the only possible inverse of x in Ais x−1, so we need to show x−1 ∈A. Since x−1 is integral over A, there are n > 0 and ai ∈A such that x−n =

∑n−1i=0 ai x−i holds in B . Multiplying

through by xn−1 yields x−1 =∑n−1

j=0 an−1− j x j , so x−1 ∈A.3 This good solution taken from .4 Here is a terrible solution I came up with myself. First we show that f (A)× g (A) is integral over the subalgebra A′ = im( f , g ) = ( f (a), g (a) :

a ∈ A= A · (1, 1). Now f (A)× g (A) is generated over A′ by (1, 0) and (0, 1), so it is finitely generated, and φ : (x, y) 7→ (b x, cy) is an A-moduleendomorphism of f (A)× g (A). (2.4) then gives us an equation of the form

∑

i≤n aiφi = 0, where an = 1; applying both sides to (1, 1) gives

∑

i≤n ai (b , c)i = (0, 0), showing (b , c) is integral over A′.Since A′′ = f (A)× g (A) is integral over A′ = im( f , g ), by (5.4) if B ×C is integral over A′′, it will also be integral over A′. It suffices by (5.3) to

show each of B×0 and 0×C is integral over A′′. We will prove it for B×0, the argument for 0×C being symmetric. So let (b , 0) ∈ B×C .As b is integral over im A⊆ B , there are ai ∈ A such that

∑

0≤i≤n ai b i = 0. Then∑

ai (b , 0)i =

0, g (a0)

∈ A′′. Setting a′i = f (ai ) for i 6= 0 anda′0 =

f (a0), 0

we have∑

ai (b , 0)i = (0, 0), so (b , 0) is integral over A′′.

75

Ex. 5.8 Chapter 5: Integral Dependence and Valuations

8. i) Let A be a subring of an integral domain B, and let C be the integral closure of A in B. Let f , g be monic polynomialsin B[x] such that f g ∈C [x]. Then f , g are in C [x].

(Note that A really plays no part: we could have started with C ⊆ B integrally closed in B , and let A= C , withintegral closure in B just C again, so without loss of generality we may take A=C .)

Let K be the field of fractions of B , let Ω be a splitting field of f g ∈ K[x]. Then in Ω[x] we have f g =∏

i (x −ξi )∏

j (x − η j ), where the ξi are roots of f and the η j are roots of g . Since each ξi and η j is a root of f g ∈ C [x],we have each ξi and η j integral over C in Ω. Recall from (5.3) that the set D of elements of Ω integral over C is aring. Since each coefficient of f =

∏

i (x − ξi ) (resp. g =∏

j (x − η j )) is a polynomial in the ξi (resp. η j ), we havef , g ∈ D[x] ∩ B[x] = (D ∩ B)[x]. But since D ∩ B consists of elements of B integral over C , and C is integrallyclosed in B , we have D ∩B =C , so f , g ∈C [x].

ii) Prove the same result without assuming that B (or A) is an integral domain.Note in particular that considering linear polynomials (x − b ), (x − c), this gives us a near-converse to [5.7]:

b + c ∈C & bc ∈C ⇐⇒ b , c ∈CWe need to see if we can alter our proof of part i) to avoid fields.5 The revised version would go as follows: let B+

be a ring containing B and such that f and g split into linear factors x−ξi and y−η j in B+[x]. These linear factorsalso divide f g , so ξi , η j are roots of f g , and so are integral over C . The coefficients of f and g , being polynomialsin the ξi and η j , are then also integral over C by (3.8). But these coefficients are in B , so by assumption also in C .

To create the extension ring B+ we need, let deg( f ) = n and deg(g ) = m, and note that we can extend B toB1 = B[x]/

f (x)

to get a larger field in which f has a root α1 = x. Then f (y) is in the kernel of the canonicalmap B1[y] B1[y]/(y − α1) since in the quotient y = α1, and f (α1) = 0. Thus f (y) is an element of the principalideal (y − α1), so there is a monic polynomial f1(y) in B1[y] such that f (y) = f1(y)(y − α1) in B1[y]. Since degreeof monic polynomials is multiplicative, we have deg( f1) = n− 1. Repeating this process, and because the degree ofthe non-linear factor decreases each time we make such an extension, we eventually get a ring B ′ = Bn over whichf splits. We can then perform a similar process for g over B ′ to get a ring (B ′)m = B+ in which both f and g splitcompletely.

9. Let A be a subring of a ring B and let C be the integral closure of A in B. Prove that C [x] is the integral closure of A[x]in B[x].

Write C ′ for the integral closure of A[x] in B[x]. Then x ∈A[x]⊆C ′ and C is integral over A, hence over A[x],so by (5.3), A[x]⊆C [x]⊆C ′.

It is now enough to show C [x] is integrally closed. Let f ∈ B[x] be such that there exist gi ∈ C [x] with gn = 1and f n +

∑n−1i=1 gi f i + g0 = 0 ∈ C [x]. Since g0 ∈ C [x], we have C [x] 3 f n +

∑n−1i=1 gi f i = f ( f n−1+

∑n−1i=1 gi f i−1)

a product of monic polynomials in B[x]. Then [5.8.ii] says that f ∈C [x].

10. A ring homomorphism f : A→ B is said to have the going-up property (resp. the going-down property) if the conclusionof the going-up theorem (5.11) (resp. the going-down theorem (5.16)) holds for B and its subring f (A).

Let f ∗ : Spec(B)→ Spec(A) be the mapping associated with f .i) Consider the following three statements:

(a) f ∗ is a closed mapping.(b) f has the going-up property.(c) Let q be any prime ideal of B and let p= qc . Then f ∗ : Spec(B/q)→ Spec(A/p) is surjective.Prove that (a) =⇒ (b) ⇐⇒ (c) (See also Chapter 6, Exercise 11.)Factorize the map canonically as f = i p for p : A f (A) and i : f (A) ,→ B the expected maps. p∗ canonically

homeomorphs Spec

f (A)

into the closed subset V

ker( f )

⊆ Spec(A) by [1.21.iv], and for each p ⊇ ker( f ), thethird isomorphism theorem (2.1.i) gives p(A)/p(p)∼=A/p, so f will satisfy any of the three properties if and only ifi does. Thus we might as well assume f : A ,→ B is an inclusion.

As far as the going-up property is concerned, by induction, it is enough to show that if p ⊆ p′ ∈ Spec(A) andq ∈ Spec(B) is such that q ∩ A = p, then there is q′ ⊇ q such that q′ ∩ A = p′. This is the same as showing eachrestriction f ∗

V (p)V (q) : V (q)→V (p) is surjective.

(a) =⇒ (b): Since V (q) ([1.15]) is closed, by assumption f ∗

V (q)

is a closed set containing f ∗(q) = p, sop ⊆ f ∗

V (q)

. (In fact, they are equal, for if p 6⊆ q′ ∩ A, then q ⊇ p is not contained in q′.) But by [1.18.ii],

V (p) = p, so f ∗

V (p)V (q) is surjective.

5 Expanded from http://pitt.edu/~yimuyin/research/AandM/exercises05.pdf

76



(b) ⇐⇒ (c): The identifications of [3.21.iii] identify f ∗

V (p)V (q) with the map f ∗ : Spec(B/q)→ Spec(A/p) induced

by f : A/p→ B/q. Then (b) holds if each f ∗

V (p)V (q) is surjective and (c) if each f ∗ is surjective, but these are essentially

the same maps.

Consider the following three statements:(a ′) f ∗ is an open mapping.(b ′) f ∗ has the going-down property.(c ′) For any prime ideal q of B, if p= qc , then f ∗ : Spec(Bq)→ Spec(Ap) is surjective.Prove that (a ′)=⇒ (b ′)⇐⇒ (c ′). (See also Chapter 7, Exercise 23).(b ′)⇐⇒ (c ′): First we should show the map of (c ′) exists. f composed with the canonical map B → Bq yields a

map f q : A→ Bq. Since each element of Sq = B\q by definition becomes a unit in Bq, and since f −1(Sq) =A\f −1(q) =A\p = Sp, each element of Sp is sent to a unit in Bq, so by (3.1) there is a unique induced map Ap→ Bq, as hoped.Call this map f q

p .By induction, the going-down property requires only that if p′ ⊆ p ∈ Spec(A) and q ∈ Spec(B) is such that

f ∗(q) = p, then there is q′ ⊆ q such that f ∗(q′) = p′. This is the same as showing each restriction f ∗

S−1q Spec(A)

S−1q Spec(B)

is

surjective, where S−1 Spec(A) is the set of primes in A not meeting S. But composing with the canonical inclusionsSpec(Ap) ,→ S−1

p Spec(A) and Spec(Bq) ,→ S−1q Spec(B) of [3.21.i], we can identify ( f q

p )∗ with this restriction.(a ′)⇐⇒ (b ′): We claim that open sets in the Zariski topology are “downward closed,” meaning p′ ⊆ p ∈ U =⇒

p′ ∈ U . Indeed, write U = X \C , C closed. Then p′ /∈ U would imply p ⊆ C , so p ⊆ C = C ; but by [1.18.ii],p=V (p′) 3 p, so p ∈C and hence p /∈U . This may be obvious, but I don’t recall having seen it proved.

Recall the notation S−1X = p ∈X : S∩p=∅ from [3.21.i] and let X = Spec(A) and Y = Spec(B). In the specialcases Sp = A\p and Sg = 1, g , g 2, . . ., write Xp = S−1

p X and Xg = S−1g X. Recall from [3.22] that Yq ≈ Spec(Bq)

is the intersection of all its basic open neighborhoods Yg (g /∈ q). Write f q : A→ Bq again for the composition of fwith the canonical map φq : B→ Bq. Then

( f q)∗

Spec(Bq) [1.21.vi]= f ∗(φ∗q

Spec(Bq) [3.21.i]= f ∗(Yq) =

⋂

g /∈q

f ∗(Yg ).

By [1.17], the Yg are open, so the Ug := f ∗(Yg ) are open. Then since q ∈ Yg and p = f ∗(q), p ∈ Ug , so all primesp′ ⊆ p are in Ug . Intersecting, Xp ⊆ f ∗q

Spec(Bq)

.Now f q factors through φp : A→ Ap as f q = f q

p φp, so taking ∗’s, by [1.21.vi] we have Xp ⊆ im(φ∗p ( fqp )∗).

But φ∗p is a homeomorphism between Spec(Ap) and the open subset Xp ⊆X by [3.21.i], so ( f qp )∗ is surjective.

11. Let f : A→ B be a flat homomorphism of rings. Then f has the going-down property.[3.18] states that f ∗ : Spec(Bq) Spec(Ap) is surjective for each q ∈ Spec(B) and p = qc . Then (c ′) =⇒ (b ′) in

[5.10] shows f has the going-down property.

12. Let G be a finite group of automorphisms of a ring A, and let AG denote the subring of G-invariants, that is of all x ∈Asuch that σ(x) = x for all σ ∈G. Prove that A is integral over AG .

The first (rather trivial) thing to do is to show AG is a ring. But indeed, by the definition of a ring homomorphismwe have σ(1) = 1 for all σ ∈G, and if a, b ∈AG , then σ(a− b ) = σ(a)+σ(−b ) = a− b and σ(ab ) = σ(a)σ(b ) = abfor all σ ∈G. Thus AG , containing 1 and being closed under subtraction and multiplication, is a subring of A.

To see AG ,→ A is integral, let x ∈ A, and let t be an indeterminate. If p :=∏

σ∈G

t − σ(x)

∈ A[t ], then eachcoefficient of p is a symmetric polynomial in the σ(x), so p ∈AG[t ]. As p is monic and 0= x− x divides p(x), wesee x is a root of p, and so x is integral over AG .

Let S be a multiplicatively closed subset of A such that σ(S) ⊆ S for all σ ∈ G, and let SG = S ∩AG . Show that theaction of G on A extends to an action on S−1A, and that (SG)−1AG ∼= (S−1A)G .

Suppose a/s = b/t ∈ S−1A. Then there is u ∈ S such that u t a = u s b in A. Applyingσ ∈G yieldsσ(u)σ(t )σ(a) =σ(u)σ(s)σ(b ) in A, meaning σ(a)/σ(s) = σ(b )/σ(t ) in S−1A. Thus if we define the action of G on S−1A by

77


σ(a/s) := σ(a)/σ(s), this definition is independent of the choice of representatives, hence well defined. It is ob-viously multiplicative, and only slightly less obviously additive:

σ

as+

bt

= σ

at + b ss t

=σ(at + b s)σ(s t )

=σ(a)σ(t )+σ(b )σ(s)

σ(s)σ(t )=σ(a)σ(s)

+σ(b )σ(t )

= σ

as

+σ

bt

.

For each a ∈ AG , we have σ(a/1) = σ(a)/1 = a/1, so the natural map AG ,→ A → S−1A factors as AG →(S−1A)G ,→ S−1A. Each element s ∈ SG becomes a unit in S−1A, hence a unit in (S−1A)G since σ(1/s) = σ(1)/σ(s) =1/s for all σ ∈G. Thus (3.1) gives a unique homomorphism χ : (SG)−1AG→ (S−1A)G taking a/s 7→ a/s .

χ is injective, for if a/s = 0 in (S−1A)G ⊆ S−1A, there is t ∈ S such that t a = 0. Then taking t ′ =∏

σ∈G σ(t ), wealso have t ′a = 0, meaning a/s = 0 already in (SG)−1AG .

On the other hand, let a/s ∈ (S−1A)G Let s ′ =∏

σ 6=idAσ(s) ∈ SG . Then since a/s and s s ′/1 are invariant, so

is their product as ′/1. Thus for every σ ∈ G we have σ(as ′)/1 = s(as ′/1) = as ′/1, so there is tσ ∈ S such thattσas ′ = tσ ·σ(as ′). Set t =

∏

τ∈G τ∏

σ∈G tσ

∈ SG . Then since tσ divides t we have σ(t as ′) = t ·σ(as ′) = t as ′, sot as ′ ∈AG . Then a/s = t s ′a/t s ′ s with t s ′a ∈AG and t s ′ s ∈ SG , so χ is surjective.

13. In the situation of Exercise 12, let p be a prime ideal of AG , and let P be the set of prime ideals of A whose contraction isp. Show that G acts transitively on P. In particular, P is finite.

Let q, q′ ∈ P . For any x ∈ q, we have∏

σ∈G σ−1(x) ∈ AG ∩ q = p. But also p = AG ∩ q′, so

∏

σ∈G σ−1(x) ∈ q′.

Thus, since q is prime, for some σ ∈G we have y = σ−1(x) ∈ q′. Then x = σ(y) ∈ σ(q′). Since x ∈ qwas arbitrary, wesee q⊆

⋃

σ∈G σ(q′). By (1.11.i), q is contained in some σ(q′). Since both q∩AG = p and σ(q′)∩AG = σ(q′)∩σ(AG) =

σ(q′∩AG) = σ(p) = p, (5.9) says we must have q= σ(q′). As q and q′ ∈ P were arbitrary, it follows that G sends anyelement of P to any other element of P , so G acts transitively. Since G is finite, and sends q to every element of P ,we have |P |= |G|/|StabG(q)| ≤ |G| finite.

14. Let A be an integrally closed domain, K its field of fractions and L a finite normal separable extension of K. Let G be theGalois group of L over K and let B be the integral closure of A in L. Show that σ(B) = B for all σ ∈G, and that A= BG .

If b ∈ B , then there it satisfies a polynomial equation∑

ai b i = 0 for some ai ∈A=AG . Applying a σ ∈G to thisequation yields 0 = σ(0) = σ

∑

ai b i

=∑

σ(ai )σ(b )i =

∑

aiσ(b )i . Thus σ(b ) satisfies a monic polynomial (the

same as b does) over A, and hence is in B . Thus σ(B)⊆ B . On the other hand, replacing σ by σ−1 in this reasoningyields σ−1(B)⊆ B , and applying σ to both sides gives B = σ

σ−1(B)

⊆ σ(B). Since σ ∈G was arbitrary, B = σ(B)for all σ ∈G.

BG = B ∩ LG = B ∩K =A, since K is the fixed field of G and A is integrally closed in K .

15. Let A, K be as in Exercise 14, let L be any finite extension field of K, and let B be the integral closure of A in L. Showthat, if p is any prime ideal of A, then the set of prime ideals q of B which contract to p is finite (in other words, thatSpec(B)→ Spec(A) has finite fibers).

Recall6 that any extension factors as a separable extension followed by a purely inseparable extension. Since aproduct of two finite numbers is finite, it suffices to show the fibers are finite for either of these kinds of extensions.

In the case of a separable extension L/K , let Ω/L/K be the least normal extension of K containing L. Write Cfor the integral closure of A in Ω; it is clearly also integral over B . If B had infinitely many primes lying over p, then(5.10) would give us at least one prime of C lying over each of those, hence infinitely many primes of C lying overp. But Ω⊇K is a finite extension, so H =Gal(Ω/K) is finite. By [5.14], A=C H , so by [5.13] there are only finitelymany primes of C lying over p.

In the case of a finite, purely inseparable extension L/K of fields of characteristic p > 0, it is well known7 thatfor each x ∈ L there is n ≥ 0 such that x pn ∈ K . If we let x1, . . . , xm generate L as a vector space over K , and letni ≥ 0 be the least exponents such that xi

pni ∈ K , then if n =maxi ni , we have for all ci ∈ K that∑m

i=1 ci xi

pn =∑m

i=1 c pn

i x pn

i ∈ K , (using the binomial theorem and the fact that p dividespn

l

for 0 < l < pn), so that Lpn ⊆ K .Write B for the integral closure of A in L, and suppose P ∈ Spec(B) lies over p ∈ Spec(A). If x ∈ B has x pn ∈ p⊆P,then as P is prime we have x ∈P; and if x ∈P, then x pn ∈P∩K = p. Thus P is determined uniquely by p, so theonly possibility for a prime of B lying over p is P := x ∈ B : x pn ∈ p. To see P really is an ideal, note that b ∈ B

6 http://planetmath.org/encyclopedia/PurelyInseparable.html7 or sometimes the definition: http://planetmath.org/encyclopedia/PurelyInseparable.html

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and x, y ∈P imply (b x)pn = b pnx pn ∈Ap= p and (x−y)pn = x pn+(−y)pn ∈ p. To see P is prime, suppose xy ∈P

but x /∈P; then x pny pn ∈ p but x pn

/∈ p, so y pn ∈ p and y ∈P.

Noether’s normalization lemma16. Let k be a field and let A 6= 0 be a finitely generated k-algebra. Then there exist elements y1, . . . , yr ∈ A which are

algebraically independent over k and such that A is integral over k[y1, . . . , yr ].We shall assume that k is infinite. (The result is still true if k is finite, but a different proof is needed.) Let x1, . . . , xn

generate A as a k-algebra. We can renumber the xi so that x1, . . . , xr are algebraically independent over k and each ofxr+1, . . . , xn is algebraic over k[x1, . . . , xr ]. Now proceed by induction on n. If n = r there is nothing to do, so supposen > r and the result true for n − 1 generators. The generator xn is algebraic over k[x1, . . . , xn−1], hence there exists apolynomial f 6= 0 in n variables such that f (x1, . . . , xn−1, xn) = 0. Let F be the homogeneous part of highest degree inf . Since k is infinite, there exist λ1, . . . , λn−1 ∈ k such that F (λ1, . . . , λn−1, 1) 6= 0. Put x ′i = xi − λi xn (1≤ i ≤ n− 1).Show that xn is integral over the ring A′ = k[x ′1, . . . , x ′n−1] and hence that A is integral over A′. Then apply the inductivehypothesis to A′ to complete the proof.

First, if the λi didn’t exist, we would have F (x1, . . . , xn−1, xn) = xdeg Fn F (x1, . . . , xn−1, 1) = 0 by homogeneity.

But k being infinite, F is zero as a function kn→ k if and only if F = 0 ∈ k[x1, . . . , xn].8

Let F =∑

I aI∏n

j=1 x i jj =

∑

I aI x inn∏n−1

j=1 (x′j +λ j xn)

i j . The coefficient of xdeg Fn in F ∈ k[x ′1, . . . , x ′n−1, xn] is

c :=∑

aIλi11 · · ·λ

in−1n−1 = F (λ1, . . . , λn−1, 1) 6= 0, so that the equation c−1 f (x1, . . . , xn−1, xn) = 0 is monic when writ-

ten in A′[xn]. Thus xn is integral over A′, and so A= k[x1, . . . , xn] is integral over A′ by (5.3). But by the inductionhypothesis, A′ is integral over some k[y1, . . . , yr ], with y1, . . . , yr algebraically independent over k, and by the tran-sitivity (5.4) of integral dependence, A is integral over k[y1, . . . , yr ].

9

From the proof [of [5.16]] it follows that y1, . . . , yr may be chosen to be linear combinations of x1, . . . , xn . This has thefollowing geometrical interpretation: if k is algebraically closed and X is an affine algebraic variety in kn with coordinatering A 6= 0, then there exists a linear subspace L of dimension r in kn and a linear mapping of kn onto L which maps Xonto L.

X ι //

ρ

kn

π

k r

A= k[x] k[t ]ι#oooo

k[y]OOπ#

OO

dd$

dd

We want the commutative diagram of regular maps on the right, with π linear. Let-ting the coordinate ring ([1.27]) of kn be k[t ], that of the affine algebraic variety X ⊆ kn

be A= k[t ]/I (X ) = k[x], and that of k r be A0 := k[y], [1.28] says that this is equivalentto demanding the diagram of k-algebra homomorphisms below it. Here ι# is the projec-tion t j 7→ x j : k[t1, . . . , tn] A. The normalization proven above gives us a candidatemap $ : A0 A, namely the k-subalgebra inclusion gotten by mapping the yi to alge-braically independent elements x ′i of A such that A is integral over A′ = k[x ′1, . . . , x ′r ].In the course of the proof above, we found that when k is infinite (which is true ifk is algebraically closed), we can take the x ′i to be k-linear combinations of the x j . Ifx ′i =

∑nj=1 ai j x j in A for ai j ∈ k, 1 ≤ i ≤ r and 1 ≤ j ≤ n, and we want ι# π# = $,

we may take π#(yi ) :=∑n

j=1 ai j t j . Since yi : k r k is the i th projection and π#(η) = η π by definition, it followsthat π should be given by (v1, . . . , vn) 7→

∑n

j=1 a1 j v j , . . . ,∑n

j=1 ar j v j ,

. This is obviously linear, and by Eq. 1.2 of[1.28], $ = ι# π# = (π ι)# = ρ# as hoped. (This map is not, as defined, to a linear subspace L ⊆ kn , but we can ifwe like pick any r linearly independent vectors wi ∈ kn and define a new map by v 7→

∑

(yi π)(v)wi .)To see that ρ is surjective, let a point of k r be given. Write it as an inclusion p : 0 → k r . It corresponds by

[1.28] to a map ψp : A0 = k[y]→ k and hence to a map A′ = k[x ′]→ k.10 Since A is integral over A′, by [5.2] this

8 Proved by induction, e.g. in Theorem 5.18 in Fields and Galois Theory, J.S. Milne, http://jmilne.org/math/CourseNotes/ft.html:if n = 1 and F 6= 0 then F has ≤ deg(F ) roots, so is not identically zero since k is infinite. Assume the result has been proved for n, let F ∈k[x1, . . . , xn+1] be zero on kn+1, and write 0 = F =

∑

Gi x in+1 for Gi ∈ k[x1, . . . , xn]. For any (a1, . . . , an) ∈ kn we have F (a1, . . . , an , xn+1) ∈

k[xn+1] identically zero by assumption, so by the n = 1 case each Gi (a1, . . . , an) = 0. Then by the induction step each Gi = 0, so F = 0.9 We include as a bonus a proof (http://ericmalm.net/ac/projects/math210b-w08/math210b-transcendence.pdf) that works

when k is finite. Again, assume that xn is algebraic over k[x1, . . . , xn−1], and say that this is witnessed by f (x1, . . . , xn) = 0. Let d > deg f , andx ′i = xi − xd i

n for i = 1, . . . , n− 1. Write xi = x ′i + xd in in f = 0. Expanding out each monomial term aI x I := aI x i1

1 · · · xinn of f in terms of xn and

the x ′i , the monomial term aI x eIn divisible only by xn will have have exponent eI = in + i1d + i2d 2 + · · ·+ in−1d n−1. By our choice of d , the

exponents eI are all distinct as we range over different aI x I , so there is no cancellation among them. One such exponent eM will be the greatest,and then we can divide through by the corresponding coefficient aM to get aM f (x ′1 + xd

n , . . . , x ′n−1 + xd n−1n , xn) = 0 monic in xn . This shows A

is integral over A′, and we conclude as before. Note that we no longer have that the xi are k-linear combinations of elements of A′, however.10 http://math.stackexchange.com/questions/24794/atiyah-macdonald-exercises-5-16-5-19

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extends to a map φb : A→ k, corresponding to an inclusion b : 0 → X . We have φb $ = ψp , so by Eq. 1.2 of[1.28] again, ρ b = p.11 We show in [8.5] that the fibers of ρare finite of bounded cardinality.

Nullstellensatz (weak form).17. Let X be an affine algebraic variety in kn , where k is an algebraically closed field, and let I (X ) be the ideal of X in the

polynomial ring k[t1, . . . , tn] Chapter 1, Exercise 27. If I (X ) 6= (1) then X is not empty.

Write k[t ] := k[t1, . . . , tn]. If I (X ) 6= (1), then A= k[t ]/I (X ) 6= 0, so by [5.16], X is carried by a linear projectiononto a linear subspace L⊆ kn . Since L is non-empty, so must be X. Let’s call this the weaker Nullstellensatz.

The name “weak Nullstellensatz” usually refers to the following related result:

Weak Nullstellensatz. If k is an algebraically closed field and aÃ k[t1, . . . , tn] is not (1), then Z(a) 6=∅.

This implies the weaker Nullstellensatz, for if X = Z(a) and I (X ) 6= (1), then since a ⊆ I Z(a) 6= (1), we havea 6= (1) and hence X = Z(a) 6=∅. The weak Nullstellensatz would also follow from the weaker Nullstellensatz if wecould prove a 6= (1) =⇒ I Z(a) 6= (1). This is an easy consequence of the strong Nullstellensatz of [7.14], but onewants to prove strong from weak, not vice versa.

Deduce that every maximal ideal in the ring k[t1, . . . , tn] is of the form (t1− a1, . . . , tn − an) where ai ∈ k.

This is also the result of [1.27], and there is a proof there. It does not seem to obviously follow from the weakerNullstellensatz above, which we are supposed to use to prove it,12 but does from the weak Nullstellensatz, which isin fact equivalent.

First assume the weak Nullstellensatz. IfmÃ k[t ] is a maximal ideal, then Z(m) 6=∅, so there exists an x ∈ Z(m),meaning m⊆ I Z(m)⊆mx ; as m is maximal, it follows m=mx . Now assume all maximal ideals of k[t ] come frompoints of kn . Any a 6= (1) is contained in some maximal idealm by (1.4), and by assumptionm=mx for some x ∈ kn ,so since mx vanishes at x by definition, x ∈ Z(a).

18. Let k be a field and let B be a finitely generated k-algebra. Suppose that B is a field. Then B is a finite algebraic extensionof k. (This is another version of Hilbert’s Nullstellensatz. The following proof is due to Zariski. For other proofs, see (5.24),(7.9).)

This is called Zariski’s Lemma, and there are other proofs at (1.27.2*), (5.24), (7.9).13 Here is a simpler proof thanthat suggested.14 Use Noether normalization ([5.16]) on the finitely generated k-algebra B : then there exist (possiblyzero) elements y1, . . . , yr ∈ B , algebraically independent over k, such that B is integral over A = k[y1, . . . , yr ]. By(5.7), B being a field implies A is a field, so there are zero y’s and thus A= k. Then B is integral over k, hence a finitealgebraic extension.

Now we proceed with the book’s intended proof.

Let x1, . . . , xn generate B as a k-algebra. The proof is by induction on n. If n = 1 the result is clearly true, so assume n > 1.

If B = k[x1] is a field, it follows x−11 ∈ B , say x−1

1 =∑n

i=0 ci x i1 for ci ∈ k. Multiplying both sides by x1 and

subtracting 1 gives 0=∑n

i=0 ci x i+11 − 1, showing x1 is algebraic over k, so B = k(x1) is a finite algebraic extension.

Let A= k[x1] and let K = k(x1) be the field of fractions of A. By the inductive hypothesis, B is a finite algebraic extensionof K, hence each of x2, . . . , xn satisfies a monic polynomial equation with coefficients in K, i.e. coefficients of the form a/b

11 This can also be verified by evaluating both sides of φb $ =ψp at each yi :

yi (p) =φb∑

ai j x j

=∑

ai j b j =π#(yi )(b ) = yi

π(b )

.

Note that kernels did not need to be mentioned here. One can also, however, prove ρ is surjective using the result of [1.27] that the maximal idealsof P (X ) are in bijection with the points of X . The regular map ρ : X → L induces by precomposition a homomorphism$ = ρ# : P (Y )→ P (X ),which in turn induces through contraction a map $∗ : Spec

P (X )

→ Spec

P (Y )

. Since A is integral over A0, by (5.8) contractions of maximalideals are maximal, so this restricts to a map eρ : Max

P (X )

→Max

P (Y )

. The identification X ↔Max

P (X )

conflates eρ with the original ρby [1.28], so it is enough to show eρ is surjective. Since A is integral over A0, this follows from (5.10) and (5.8).

12 This has caused me some consternation; see the discussion at http://math.stackexchange.com/questions/24794/atiyah-macdonald-exercises-5-16-5-19.

13 The original proof, from Oscar Zariski, “A new proof of Hilbert’s Nullstellensatz”, Bull. Amer. Math. Soc. Volume 53, Number 4 (1947),362–368, can be found online at http://projecteuclid.org/DPubS?verb=Display&version=1.0&service=UI&handle=euclid.bams/1183510605.

14 http://www.math.lsa.umich.edu/~hochster/615W10/supNoeth.pdf

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http://projecteuclid.org/DPubS?verb=Display&version=1.0&service=UI&handle=euclid.bams/1183510605

http://www.math.lsa.umich.edu/~hochster/615W10/supNoeth.pdf


where a and b are in A. If f is the product of the denominators of all these coefficients, then each of x2, . . . , xn is integralover Af .

Write ai/bi for the coefficients, with ai , bi ∈A. If f =∏

bi , then ai/bi = ai

∏

j 6=i b j

/ f ∈Af .

Hence B and therefore K is integral over Af .

By (5.3), B =A[x2, . . . , xn] is integral over Af , so since K ⊆ B , we see K is also integral over Af .

Suppose x1 is transcendental over k. Then A is integrally closed, because it is a unique factorization domain.

First we show a UFD A is integrally closed. Suppose an element of its field of fractions is integral over A. We canwrite it in least terms as a/b , since A has unique factorization. Then (a)+(b ) = (1) in A, so (an)+(b ) = (1) for all n by(1.16). We have an equation 0= (a/b )n +

∑n−1i=0 ci (a/b )i , and multiplying by b n gives an =−

∑n−1i=0 ci a

i b n−i ∈ (b ).But then (1) = (an)+ (b ) = (b ), so b is a unit and a/b ∈A.

Now, if x1 is transcendental over k, then A= k[x1] has a division algorithm, so it is a PID and hence a UFD.

Hence Af is integrally closed (5.12), and therefore Af =K, which is clearly absurd.

(5.12) says Af is the integral closure of Af in K f∼= K. But in the previous paragraph we showed K was integral

over Af , so K =Af . To see this is impossible, see (5.18.1*) below.15

Hence x1 is algebraic over k, hence K (and therefore B) is a finite extension of k.

We take this opportunity to prove a more general result, the Zariski–Goldman–Krull theorem.16

Definition. A Goldman domain is a domain A containing some element a such that the localization Aa is a field.

Note that then Aa is the field of fractions of A. Note also that an iterated localization

(Aa)···

z =Aa···z , so we canequivalently say a Goldman domain is a domain A whose field of fractions is a finitely generated A-algebra.

Lemma 5.18.1*. No polynomial ring A[x] is a Goldman domain.

Proof. Assume A is a domain: if not, neither would A[x] be Let K be the field of fractions of A. If A[x] were aGoldman domain, then so would K[x] be, since K[x] = K ·A[x] and K(x) = K ·A(x). Cribbing from Euclid, notethat given any finite list of irreducible polynomials pi ∈ K[x], none divides 1+

∏

pi , so there are infinitely manyirreducibles in K[x]. Since K[x] is a UFD, there are then for any f ∈A[x] irreducible p not dividing any power f n ,so that 1/p /∈K[x] f .17

Corollary 5.18.2*. If a field L contains a subfield K and there exist elementsβ ∈ L and b ∈K[β] such that K[β]b = L,then β is integral over K; in this case, K[β] is a field, so b−1 ∈K[β] as well and L=K[β].

Lemma 5.18.3*. Suppose A⊆A[β] = B ⊆ Bb = L, where L is a field. Then there exists a ∈A such that Aa is a field andL is a finite extension of Aa .

Proof. Write K for the field of fractions of A. Since L = Bb = A[β]b , we see L = K[β]b as well, so by (5.18.2*), βis integral over K and L = K[β]. Thus [L : K] is finite and b ∈ L is integral over K . Multiplying denominators inthe equations witnessing integrality of b and β over K , we obtain an a ∈ A such that b and β are integral over Aaand hence the field L=Aa[β]b is integral over Aa . But then, by (5.7) or [5.5.i], Aa must be a field, and hence, beingintermediate between A and its field of fractions, so must itself be K .

Zariski–Goldman–Krull Theorem. If a field L is a finitely generated algebra over a subring A, then there exists a ∈ Asuch that Aa is a field and L is a finite extension field of Aa .

15 Alternately, find a non-zero proper ideal of Af . By (3.11.iv), any ideal of A not meeting S f = 1, f , f 2, . . . yields a proper ideal of Af . But,for example, (1− f ) does not meet S f , by unique factorization in k[ f ], so (1− f )ÃAf is a non-zero proper ideal.

16 This sequence of results is a mild reformulation of the proof given by Daniel J. Bernstein at http://cr.yp.to/zgk.html.17 This proof is from Proposition 12.5 of Pete L. Clark’s http://math.uga.edu/~pete/integral.pdf.Here is an alternate proof taken from Richard G. Swan, “On Munshi’s proof of the Nullstellensatz,” at http://www.math.uchicago.edu/

~swan/nss.pdf. Suppose for a contradiction there exists f ∈ A[x] such that A[x] f is a field. Then f /∈ A, for otherwise we would haveA[x] f =Af [x] a polynomial ring, so deg f ≥ 1, and in particular 1− f 6= 0. Since (1− f )−1 = g/ f n for some g ∈ A[x], clearing denomi-nators yields f n = (1− f )g in A[x]. Modulo 1− f , we have 1≡ f , so 1≡ f n ≡ (1− f )g ≡ 0, meaning 1− f is a unit of A[x]; but deg(1− f )≥ 1and A contains no nonzero nilpotents, so this is impossible by [1.2.i].

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Proof. The proof proceeds by induction on the number n of generators for L over A. For n = 0, the result is trivial,since L = A= A1. Assume the result proved for n generators and let L′ = A[α1, . . . , αn+1] for some α j ∈ L′. WriteB = A[α1] and L for its field of fractions. The induction hypothesis, applied to the extension B ⊆ L′, yields b ∈ Bsuch that Bb (= L) is a field and [L′ : L] is finite; and (5.18.3*), applied to A⊆ L, yields a ∈ A such that Aa is a fieldand [L : Aa] is finite. Then [L′ : Aa] = [L

′ : L][L : Aa] is finite, concluding the induction.

Zariski’s Lemma is an immediate corollary. We will also meet Jacobson rings in [5.23], and show in (5.23.5*)that a ring is Jacobson if and only if its every quotient Goldman domain is a field.

Corollary 5.18.4*. If a field L is a finitely generated algebra over a quotient domain B of a Jacobson ring A, then B is afield and L is a finite extension of B.

Proof. By the Zariski–Goldman–Krull Theorem, B is a Goldman domain and L is a finite extension of its field offractions. Since A is Jacobson, B is a field by (5.23.5*).

Note that this is also the direction i) =⇒ ii) of [5.25].

Generalized Nullstellensatz. If A is a Jacobson ring and C a finitely generated A-algebra, then C is a Jacobson ring. IfmÃC is a maximal ideal of C , then mc is a maximal ideal of A and C/m is a finite extension field of A/mc .18

Proof. The first statement is (ii) from [5.24]. For the second, write A′ = imA⊆C and p=m∩A′. Then L=C/m isa field finitely generated over the domain B =A′/p, so by (5.18.4*), L is finite over B and B is a field. This means p ismaximal. By the correspondence (1.1) applied to AA′, it follows mc ÃA is maximal and A/mc ∼=A′/p= B .

Note how the second clause generalizes (1.27.3*): the codomain is now allowed to be any Jacobson ring finitelygenerated over the domain.

19. Deduce the result of Exercise 17 from Exercise 18.

Let k be an algebraically closed field. We prove (1.27.4*) from [1.27], namely that all maximal ideals m of k[t ] :=k[t1, . . . , tn] come from points; the other results then follow as explained in [5.17]. B = k[t ]/m is a field finitelygenerated as a k-algebra, so by [5.18] it is a finite extension of k. Since k is algebraically closed, this gives a k-algebraisomorphism φ : B

∼−→ k. If ti 7→ xi under the composition k[t ] B∼−→ k, then ti − xi ∈ m, so mx ⊆ m. As mx is

maximal, the two are equal.

20. Let A be a subring of an integral domain B such that B is finitely generated over A. Show that there exists s 6= 0 in A andelements y1, . . . , yn in B, algebraically independent over A and such that Bs is integral over B ′s , where B ′ =A[y1, . . . , yn].

Can we invoke ZGK?Since B is an integral domain, so must A be. Let S =A\0, so that k = S−1A is a field. Since B is finitely generated

over A, S−1B is finitely generated over k. By Noether normalization ([5.16]), there exist elements y1/s1, . . . , yn/snof S−1B , with yi ∈ B and si ∈ S, which are algebraically independent over k and such that S−1B is integral overC = k[y1/s1, . . . , yn/sn]. It follows that the y j are also algebraically independent over A. Let x1, . . . , xr generate Bover A; then the xi/1 generate S−1B over k, and a fortiori over C . Since S−1B is integral over C , each xi/1 satisfiesa monic polynomial pi (x) =

∑

ci , j (xi/1)j in C [x]. Let s ∈ S be so large that s ci , j ∈ B ′ for all i , j (multiply all the

denominators). Then pi (x) ∈ B ′s [x] for each i , so each xi/1 is integral over B ′s . Since Bs = B ′s [x1/1, . . . , xr/1], we seefrom (5.3) that Bs is integral over B ′s .

21. Let A, B be as in Exercise 20. Show that there exists s 6= 0 in A such that, if Ω is an algebraically closed field and f : A→Ωis a homomorphism for which f (s) 6= 0, then f can be extended to a homomorphism B→Ω.

Recall s ∈ A\0 from the previous proof, and suppose f (s) 6= 0. Then by (3.1) f extends uniquely to a mapfs : As → Ω. Next, B ′s = As [y1, . . . , yn] is a polynomial ring over As , so we may pick any ωi ∈ Ω and extend fs tof ′s : B ′s → Ω by yi 7→ωi . As Bs is integral over B ′s , by [5.2] we may extend f ′s to gs : Bs → Ω. Recalling the canonicalmap φs : B→ Bs , define g = gs φs : B→Ω. By our definitions, g |A= f .

18 This shows up for instance as Theorem 4.19 in Eisenbud.

82


22. Let A, B be as in Exercise 20. If the Jacobson radical of A is zero, then so is the Jacobson radical of B.

Let 0 6= b ∈ B . Since the Jacobson radical is defined as the intersection of the maximal ideals, we want to finda maximal ideal n of B not containing b . This is the same as finding a map g : B B/n to a field with g (s) 6= 0.As every field has an algebraic closure ([1.13]), it will suffice (WHY IS THIS ENOUGH? WHAT GUARANTEESTHE KERNEL IS MAXIMAL?) to find an algebraically closed field Ω and a map g : B → Ω such that g (b ) 6= 0.If this map exists, by (3.1) it will have a unique extension gb : Bb → Ω. Now Bb = B[1/b ] is finitely generated asa B -algebra, and B is finitely generated as an A-algebra, so Bb is finitely generated as an A-algebra. Let s ∈ A\0,as in the previous problems, correspond to the extension Bb ⊇ A. Since the Jacobson radical of A is 0, there is ahomomorphism f : A → Ω with f (s) 6= 0, and by [5.21] it extends to a homomorphism gb : Bb → Ω. Then therestriction g = gb |B has g (b ) 6= 0 and is the map we were after.

23. Let A be a ring. Show that the following are equivalent:i) Every prime ideal in Ais an intersection of maximal ideals.ii) In every homomorphic image of Athe nilradical is equal to the Jacobson radical.iii) Every prime ideal in Awhich is not maximal is equal to the intersection of the prime ideals which contain it strictly.

i) =⇒ ii): Let a Ã A and write M (a) = V (a)∩Max(A) for the set of maximal ideals containing a. The radicalr (a) =

⋂

V (a), and since each p ∈ V (a) is by assumption i) equal to⋂

M (p), we also have r (a) =⋂

M (a). In thequotient A/a we then have N=R by the correspondence (1.1).

ii) =⇒ iii): Let p ∈ Spec(A) not be maximal. Then (0)ÃA/p is not maximal by the correspondence (1.1). SinceA/p is an integral domain, (0) is the nilradical, which by assumption ii) equals the Jacobson radical, the intersectionof Max(A/p). Then (0) is a fortiori the intersection of Spec(A/p)\

(0)

. That means that upstairs in A, p is theintersection of V (p)\p, the set of primes that strictly contain p.

iii) =⇒ i): Two failed approaches are footnoted here.19 These failing, we follow the book’s hint. Assume pÃA isa prime ideal that is not an intersection of maximal ideals, so that in B =A/p, the trivial ideal (0) is not an intersectionof maximal ideals. In particular, the Jacobson radical R(B) 6= (0), so there exists a non-zero f ∈R(B). Let Yf be theset of primes of B not meeting S f = 1, f , f 2, . . .. Yf is not empty, as it contains (0). By (1.3), B f contains a maximalideal B f q, and by the correspondence (3.11.iv) its contraction q is a prime ideal maximal with respect to not meetingS f , hence a maximal element of Yf . But as assumption iii) continues to hold in B (whose prime and maximal idealsare by (1.1) images of those in A), it follows that q is an intersection of prime ideals containing f . Then q contains fas well, which is a contradiction.

A ring A with the three equivalent properties above is called a Jacobson ring.20

Lemma 5.23.1*. A homomorphic image of a Jacobson ring is Jacobson.

Proof. Let φ : A B be a ring surjection, and q ∈ Spec(B). If A is Jacobson, we can write qc =⋂

mα for somemα ∈Max(A), and then q= qc e =

⋂

meα; but by (1.1), the me

α Ã B are maximal, so by condition i), B is Jacobson.

A rephrasing of condition i) is that A is a Jacobson ring just if for every quotient domain B we have R(B) = 0.We now relate Jacobson rings to the Goldman domains introduced in [5.18].

Definition. A prime ideal p ∈ Spec(A) is a Goldman ideal if A/p is a Goldman domain. For a Ã A, let G(a) denotethe set of Goldman ideals p containing a.

Lemma 5.23.2*. For any ring A and ideal aÃA, we have r (a) =⋂

G(a).21

19 One approach would use an induction argument on the length of chains of primes containing p ∈ Spec(A). Suppose that each chain P ⊆V (p)\p ([1.15]) of primes strictly containing p is well-ordered by⊇. Then assign to each P an ordinal α(P ) describing its order-type, and defineα(p) = supα(P ) as P ranges over chains in V (p)\p. If α(m) = 0, then m is maximal and trivially an intersection of maximal ideals. Supposeα(q) = β and each p with α(p) < β is an intersection of maximal ideals. Then by iii) q is an intersection of primes p with α(p) < β and so isitself an intersection of maximal ideals. This attempt fails because the relation ⊇ (resp. ⊆) on Spec(A) is not in general well-founded if A is notNoetherian (resp. Artinian). In the ring A= k[x1, xn , . . .] of example 6) on p. 75, if we take an = (x1, . . . , xn) and bn = (xn , xn+1, . . .), then anis an infinite ascending series of primes of A and bn an infinite decreasing series of primes.

The other approach was to let Σ be the set of prime ideals that are not intersections of maximal ideals and show Σ=∅. If p is maximal in Σ,then all primes containing it are intersections of maximal ideals, so by iii), p is itself an intersection of maximal ideals; thus Σ cannot have anymaximal elements. If the assumption that Σ is non-empty leads to a proof Σ has a maximal element, then we will have shown Σ = ∅. I wantedto assume Σ was nonempty and then use Zorn’s Lemma to show Σ has maximal elements; it’s not clear, however, that a chain in Σ has an upperbound in Σ.

20 More on these rings can be found in Matthew Emerton’s notes http://www.math.uchicago.edu/~emerton/pdffiles/jacobson.pdfand Pete L. Clark’s notes http://math.uga.edu/~pete/integral.pdf.

21 This statement and proof are from Proposition 12.9 in Pete L. Clark’s notes http://math.uga.edu/~pete/integral.pdf.

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http://www.math.uchicago.edu/~emerton/pdffiles/jacobson.pdf




Proof. Substituting A/a for a and using the correspondence (1.1), it is enough to show N =⋂

G(0). The contain-ment N ⊆

⋂

G(0) follows by (1.8). For the other direction, suppose a ∈ A\N. Then Aa 6= 0, and so by (3.11.iv), amaximal ideal of Aa contracts to a prime ideal pÃA maximal with respect to the property of not containing a. Sinceevery larger prime contains a, by (1.1), every nonzero prime of the domain A/p contains a. Therefore, by (3.11.iv)again, no nonzero prime survives in (A/p)a , which then must be a field. It follows that A/p is a Goldman domain,so that p is a Goldman ideal not containing a.

Lemma 5.23.3*. If a ring A is such that every Goldman ideal is maximal, then A is Jacobson.

Proof. Each prime p= r (p)(5.23.2*)=

⋂

G(p) =⋂

M (p), by assumption, so satisfying condition i).

Lemma 5.23.4*. If a Goldman domain A has zero Jacobson radical, then it is a field.

Proof. Suppose a domain A is not a field, but there exists an element a ∈ A such that Aa is a field. Then for everyelement b of every maximal ideal m of A, there exists an inverse c/an in Aa , so that b · (c/an) = 1, or b c = an . Thenan ∈m, so a ∈m, and hence a ∈R(A).

Proposition 5.23.5*. A ring A is Jacobson if and only if every quotient which is a Goldman domain is a field.

Proof. =⇒: If pÃA is a Goldman ideal, then A/p is a domain, so 0=N(A/p) =R(A/p) by condition ii) for Jacobsonrings. By (5.23.4*), A/p is a field.⇐=: This is (5.23.3*).

24. Let A be a Jacobson ring (Exercise 23) and B an A-algebra. Show that if B is either (i) integral over A or (ii) finitely generatedas an A-algebra, then B is Jacobson.

(ii): Let q ∈ Spec(B) and p = qc . Then B ′ = B/q is an integral domain finitely generated over A′ = A/p. SinceA was Jacobson, R(A′) = N(A′) = (0), so by [5.22], the Jacobson radical R(B ′) = (0). This shows that in q is theintersection of the maximal ideals of B containing it.

(i): Let q ∈ Spec(B) and b ∈ b, the intersection of the maximal ideals containing q. Write f : A → B for thehomomorphism making B an A-algebra. Then B is integral over the subring f (A)[b ]. By (5.23.1*), f (A) is Jacobson,and by (ii) above, so is f (A)[b ]. Thus we may assume A⊆ B and a := b ∈A∩b. Now B ′ = B/q is an integral domain,integral over A′ = A/qc by (5.6.i), and b/q = R(B ′). Since A was Jacobson, by (5.23.1*) again, A′ is Jacobson, soR(A′) = 0. But by [5.5.ii], R(B ′)∩A′ =R(A′), so b ∈R(A′) = 0, meaning b ∈ q.

In particular, every finitely generated ring, and every finitely generated algebra over a field, is a Jacobson ring

A ring is finitely generated if it is finitely generated as a Z-algebra, so by (ii) above it will suffice to show that Zand all fields are Jacobson. But the prime ideals of Z are all either maximal themselves or (0) =R, so Z is Jacobsonby condition i) of [5.23], and similarly fields are Jacobson because their prime ideal is (0).

25. Let A be a ring. Show that the following are equivalent:i) A is a Jacobson ring;ii) Every finitely generated A-algebra B which is a field is finite over A.

i) =⇒ ii): Write A′ for the image of the map A→ B ; as a quotient of A, it is Jacobson, and as a subset of B , itis an integral domain; thus R(A′) =N(A′) = 0. Find an element s 6= 0 in A′ as in [5.20], [5.21]. Then there is somemaximal ideal m of A′ not containing s . If we let k =A′/m andΩ be the algebraic closure of k, then the compositionf : A′ k ,→ Ω doesn’t send s to 0, so by the assumption of [5.21], f extends to a homomorphism g : B → Ω. Asa map of fields, g is injective, so B ∼= g (B). Since B is finitely generated over A′, say B = A′[y1, . . . , yn], the imageg (B) is generated over k by the g (yi ). But the g (yi ) are algebraic over k, so g (B) is a finitely generated k-module,thus a finitely generated A′-module, and finally a finitely generated A-module. (Question: unless A′ is already a field,doesn’t the fact that g : B→Ω extends A′→A′/m= k contradict g ’s being an injection?)

ii) =⇒ i): Let pÃA be a prime ideal, not maximal, and consider A′ =A/p. We want to show the intersection ofprimes strictly containing p in A is p; downstairs in A′, we want to show the intersection of the non-zero primes is0. Equivalently, for every nonzero s ∈A′, Ss = 1, s , s2, . . .misses some non-zero prime ideal. Now A′s is a finitelygenerated A′-algebra. If it is a field, then by assumption, it is finite over A, hence integral over A′, and (5.7) says thatA′ is a field, so p is maximal. So it is not a field, and it has a nonzero maximal ideal qs , whose contraction to A′ is aprime q not meeting Ss by (3.11.iv).

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Note that the direction i) =⇒ ii) shows that Jacobson rings A are the furthest generalization of fields k for whichZariski’s Lemma ((1.27.2*), (5.24), [5.18], (7.9)) still holds.

26. Let X be a topological space. A subset of X is locally closed if it is the intersection of an open set and a closed set, orequivalently if it is open in its closure.

We prove these conditions are equivalent. Supposing C is closed and U is open in X, we want to show C ∩U isopen in its closure. LetC be the collection of all closed sets of X containing U ; then every closed set of X containingC ∩U contains some member of C ′ = K ∩C : K ∈C , so C ∩U =

⋂

C ′ =C ∩U . Then C ∩U = U ∩ (C ∩U )is indeed open in C ∩U .

On the other hand, if S ⊆X be given such that S is open in its closure C = S, then there is by definition an openU ⊆X such that S =U ∩C .

The following conditions on a subset X0 of X are equivalent:(1) Every non-empty locally closed subset of X meets X0;(2) For every closed set E in X we have E ∩X0 = E;(3) The mapping U 7→U ∩X0 of the collection of open sets of X onto the collection of open sets of X0 is bijective.

(1) =⇒ (2): If E is closed, x ∈ E , and U is any neighborhood of x, then U ∩ E is locally closed, so by (1),U ∩E ∩X0 6=∅. Thus every neighborhood of x meets E ∩X0, so x ∈ E ∩X0. Thus E ⊆ E ∩X0. On the other hand,E ∩X0 ⊆ E = E .

(2) =⇒ (3): By the definition of the subspace topology, the mapping U 7→U ∩X0 is surjective from the topologyof X to that of X0. To see injectivity, suppose U ∩X0 = V ∩X0; taking complements in X0, we get (X \U )∩X0 =(X \V ) ∩ X0. Taking closures gives (X \U )∩X0 = (X \V )∩X0. Since X \U and X \V are closed in X, (2) givesX \U =X \V ; and taking complements finally shows U =V .

(3) =⇒ (1): A locally closed subset of X is of the form C ∩U = U \V for C = X \V closed and U open. IfC ∩U doesn’t meet X0, then X0 ∩U ∩C = ∅, so X0 ∩U ⊆ X \C = V . Intersecting both sides with X0 ∩U givesX0 ∩U ⊆ X0 ∩U ∩V , but on the other hand since U ∩V ⊆ U , intersecting with X0 gives X0 ∩U ∩V ⊆ X0 ∩U .Thus U and U ∩V have the same image under the map of (3), so by assumption, U = U ∩V , or U ⊆ V . ThenC ∩U =U \V =∅.

A subset X0 satisfying these conditions is said to be very dense in X.If A is a ring, show that the following are equivalent:

i) A is a Jacobson ring;ii) The set of maximal ideals of A is very dense in Spec(A);iii) Every locally closed subset of Spec(A) consisting of a single point is closed.

i) ⇐⇒ ii): Let a Ã A be an arbitrary ideal, so that V (a) ⊆ X = Spec(A) ([1.15]) is an arbitrary closed subset,and let b=

⋂

V (a)∩Max(A)

be the intersection of all maximal ideals containing a. Then by Eq. 1.1 from [1.18.i],the closure of V (a)∩Max(A) is V (b). Since a is a subset of each prime in the set V (a)∩Max(A), we have a⊆ b, soV (b) ⊆ V (a). By (2) above, Max(A) is very dense in X just if V (a) ⊆ V (b) for all a, so every prime containing acontains b. This happens just if in in every quotient A/a, every prime contains the Jacobson radical, so the Jacobsonradical and the nilradical are equal. By [5.23.ii], this happens if and only if A is a Jacobson ring.

i) ⇐⇒ iii): A locally closed subset S ⊆X can be written as S =V (a)∩U with U open. If we write U =X\V (b),then S =V (a)\V (b). If S is a singleton, there is exactly one prime ideal p containing a that does not contain b. Writec= r (p+ b). By [1.15.i,iii], V (c) =V (p∪ b) =V (p)∩V (b)⊆V (p). Then S = p=V (a)\V (b) =V (p)\V (c), soall primes strictly containing p contain c, and c=

⋂

(V (p)\p) is strictly bigger than p. Now [1.18.i] says that eachlocally closed singleton S = p is closed if and only if each such p is maximal (and c= (1), V (c) =∅) if and only ifthe only primes p that are not intersections of larger primes are maximal; but this last condition says A is Jacobson,by [5.23.iii].

Valuation rings and valuations27. Let A, B be two local rings. B is said to dominate A if A is a subring of B and the maximal ideal m of A is contained in

the maximal ideal n of B (or, equivalently, if m= n∩A). Let K be a field and let Σ be the set of all local subrings of K. IfΣ is ordered by the relation of domination, show that Σ has maximal elements and that A∈Σ is maximal if and only ifA is a valuation ring of K.

Given a chain (Aα, mα) in this ordering, the union A=⋃

Aα is a subring of K and m=⋃

mα is an ideal of A. Ifkα = Aα/mα are the residue fields, then we have a natural embedding kα ,→ kβ for α ≤ β, and if α ≤ β ≤ γ , then

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the canonical embedding kα ,→ kγ is the composition kα ,→ kβ ,→ kγ . Thus the chain defines a direct system of fieldhomomorphisms with direct limit k. Since the diagrams of short exact sequences

0 // mα // _

Aα _

// // kα //

0

0 // mβ // Aβ // // kβ // 0

are commutative, they give rise ([2.18,19]) to a short exact sequence 0→m ,→A k→ 0 of direct limits, showingm is a maximal ideal of A. Thus each chain has an upper bound, so Zorn’s Lemma gives maximal elements.22

Let (A, m) be a maximal element and Ω the algebraic closure of A/m. Then if f : A A/m ,→ Ω is the expectedmap, (A, f ) is an element of the set called Σ on p. 65. If we have (A, f ) ≤ (A, f ′) in this order, then A ⊆ A′ andf ′|A= f , so ker( f ′)∩A= ker( f ) =m. By maximality of (A, m), this means A′ =A and f ′ = f , so (A, f ) is a maximalelement in its ordering, and (5.21) says that (A, m) is a valuation ring of K .

If on the other hand (A, m) is a valuation ring dominated by (B , n), we show they are equal. By (5.18.ii), B is avaluation ring as well. Write m− =m\0=

K\A−1 and n− = n\0=

K\B−1. As (B , n) dominates (A, m), we

have m− ⊆ n− and m−1− ⊆ n−1

− . By definition, B\A⊆ B , but by what we’ve shown, B\A⊆K\A=m−1− ⊆ n−1

− =K\B ,so we conclude B\A=∅ and B =A.

28. Let A be an integral domain, K its field of fractions. Show that the following are equivalent:(1) A is a valuation ring of K;(2) If a, b are any two ideals of A, then either a⊆ b or b⊆ a.

(1) =⇒ (2): Suppose a 6⊆ b, so there is a ∈ a\b. Obviously a 6= 0. If b= 0, then b⊆ a; otherwise there is a nonzerob ∈ b. Then a/b 6= 0, so either a/b ∈ A or b/a ∈ A. It it were the former, we would have a = (a/b )b ∈ Ab = b,contrary to assumption, so b/a ∈A. Thus b = (b/a)a ∈Aa= a. Since b ∈ b\0 was arbitrary, b⊆ a.

(2) =⇒ (1): Let a/b ∈ K× for a, b ∈ A and b 6= 0. Then a ∈ (a) ⊆ (b ) or b ∈ (b ) ⊆ (a) in A. In the formercase, write a = x b with x ∈ A; then a/b = x b/b = x ∈ A. In the latter case, write b = ya with y ∈ A; thena/b = a/ya = 1/y, so (a/b )−1 = y ∈A.

Deduce that if A is a valuation ring and p is a prime ideal of A, then Ap and A/p are valuation rings of their fields offractions.

Since the containment relation on ideals of Ap or A/p is inherited from A, and both rings are still integral domains,they are also valuation rings.

29. Let A be a valuation ring of a field K. Show that every subring of K which contains A is a local ring of A.Let A⊆ B ⊆ K be rings. By (5.18.ii), B is a valuation ring, so by (5.18.i) it is local with maximal ideal p. If m is

the maximal ideal of A, we have p ⊆ m, for if 0 6= x ∈ B with x−1 /∈ B , since A⊆ B we have x−1 /∈ A, and as A is avaluation ring, x ∈A. Then p= p∩A is a prime ideal of A. We claim B =Ap.

Slightly contrary to our usual notation, write S−1 = x ∈K : x−1 ∈ S for S ⊆K . Since for each x ∈K× we havethat x ∈A or x−1 ∈A, or both, and similarly for B , we get decompositions K =mqA×qm−1 and K = pqB×qp−1,as in the figure below.

m A× m−1

p B× p−1

Since obviously A⊆Ap, it remains to show B\A⊆Ap, but it is evident from the figure that B\A⊆ (m\p)−1 (actually,they are equal). To prove it without reference to the figure, note that since A=A×∪m⊆ B , the first decompositionimplies B\A⊆m−1, and since B ∩ p−1 =∅, we have B\A⊆m−1\p−1 = (m\p)−1 ⊆ (A\p)−1 ⊆Ap.

Note that this result does not contradict [5.27], since p⊆m and therefore B does not dominate A.

30. Let A be a valuation ring of a field K. The group U of units of A is a subgroup of the multiplicative group K× of K.Let Γ =K×/U . If ξ , η ∈ Γ are represented by x, y ∈K, define ξ ≥ η to mean xy−1 ∈A. Show that this defines a total

ordering on Γ which is compatible with the group structure (i.e., ξ ≥ η =⇒ ξω ≥ ηω for all ω ∈ Γ ). In other words, Γis a totally ordered abelian group. It is called the value group of A.

22 My first inclination was to try to use the theorem as suggested, but the set Σ on p. 65 depends on choosing an algebraically closed field Ωand it’s not immediate apparent what field to choose to be codomain for an entire chain. Moreover, depending what valuation ring one chooses,the target field changes. For example, for each nonzero (p) ∈ Spec(Z), the residue field of Z(p) (Q is Fp .

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Well-definedness: Let x, x ′ ∈K× represent ξ ∈ Γ and y, y ′ ∈K× represent η ∈ Γ . We show that the relation ξ ≥ ηis independent of the representatives chosen. ξ = U x = U x ′, so U x ′x−1 = U , meaning x ′x−1 ∈ U , and similarlyy(y ′)−1 ∈ U . The x, y version of ξ ≥ η gives xy−1 ∈ A. Then x ′(y ′)−1 = [x ′x−1][xy−1][y(y ′)−1] ∈ U−1AU = A,giving the x ′, y ′ version of ξ ≥ η.

Reflexivity: If x ∈K× represents ξ ∈ Γ , then x x−1 = 1 ∈A, showing ξ ≥ ξ .Antisymmetry: Let x, y ∈ K× respectively represent ξ , η ∈ Γ . If ξ ≥ η and η≥ ξ , then xy−1 ∈A and y x−1 ∈A.

Since (xy−1)(y x−1) = 1, this shows xy−1 ∈U , so ξ =U x =U y = η.Transitivity: Let x, y, z ∈K× respectively represent ξ , η, ζ ∈ Γ . If ξ ≥ η and η≥ ζ , then xy−1 ∈A and y z−1 ∈A,

so multiplying them, x z−1 = (xy−1)(y z−1) ∈A, and ξ ≥ ζ .Compatibility: Let x, y, w ∈ K× respectively represent ξ , η,ω ∈ Γ . If ξ ≥ η, then xy−1 ∈ A. But xy−1 =

x(ww−1)y−1 = (xw)(yw)−1, showing ξω ≥ ηω.

Let v : K×→ Γ be the canonical homomorphism. Show that v(x + y)≥min

v(x), v(y)

for all x, y ∈K×.Without loss of generality, let v(x)≥ v(y), so that xy−1 ∈A. Then A3 xy−1+1= (x+y)y−1, so v(x+y)≥ v(y).

Note also that v(xy) = xyU = xU · yU = v(x)v(y), so v is a valuation with values in Γ , in the terminology of thefollowing exercise.

31. Conversely, let Γ be a totally ordered abelian group (written additively), and K a field. A valuation of K with values inΓ is a mapping v : K×→ Γ such that(1) v(xy) = v(x)+ v(y),(2) v(x + y)≥min

v(x), v(y)

,for all x, y ∈K×. Show that the set of elements x ∈K× such that v(x)≥ 0 is a valuation ring of K. This ring is called thevaluation ring of v, and the subgroup v(K×) of Γ is the value group of v.

The book’s statement needs to be corrected mildly: the ring surely needs 0 ∈ K as well. The traditional way tofix this is to add a new element∞ to Γ , and let ∆= Γ ∪ ∞ be a monoid with subgroup Γ such that ξ +∞=∞for all ξ ∈∆.23 One extends the order on Γ by∞≥ ξ for all ξ ∈∆ and defines v(0) :=∞. This extended valuationv satisfies (1) since v(0 · x) = v(0) =∞=∞+ v(x) = v(0)+ v(x) and (2) since v(0+ x) = v(x) =min

∞, v(x)

=min

v(0), v(x)

.Now let A := x ∈ K : v(x) ≥ 0; we verify A is a valuation ring. We also verify m = x ∈ K : v(x) > 0 is the

unique maximal ideal of A. Since associativity, commutativity, distributivity, and identities are inherited from K , wehave only to check closure properties of A.

• 0 ∈A: Note v(0) =∞≥ 0.

• 1 ∈A: Note v(1) = v(1 · 1) = v(1)+ v(1), so subtracting off v(1) gives v(1) = 0.

• −1 ∈A: Note 0= v(1) = v(−1 ·−1) = v(−1)+ v(−1) = 2v(−1). If v(−1)> 0, then 0= 2v(−1)> 0, which isfalse; similarly, v(−1)< 0 would imply 0= 2v(−1)< 0, which is false; so v(−1) = 0.

• x ∈A =⇒ −x ∈A: If x ∈A, then v(−x) = v(−1 · x) = v(−1)+ v(x) = 0+ v(x) = v(x)≥ 0.

• x, y ∈A =⇒ x + y ∈A: If v(x), v(y)≥ 0, then v(x + y)≥min

v(x), v(y)

≥ 0.

• x, y ∈A =⇒ xy ∈A: If v(x), v(y)≥ 0, then v(xy) = v(x)+ v(y)≥ 0.

• v(x−1) =−v(x): 0= v(1) = v(x x−1) = v(x)+ v(x−1); subtract v(x) from both sides.

• x /∈A =⇒ x−1 ∈A: If x /∈A, then v(x)< 0, so v(x−1)≥ 0 and x−1 ∈A.

• x ∈A\m =⇒ x ∈A×: If v(x) = 0, then v(x−1) = 0, so x−1 ∈A and x ∈A×.

Thus the concepts of valuation ring and valuation are essentially equivalent.To prove this statement, we should verify that these correspondences are inverse. Let A be a valuation ring of

a field K , and let v : K× K×/A× =: Γ be the canonical map. [5.30] shows it is a valuation. Its valuation ring isA′ := 0 ∪ x ∈ K× : v(x)≥ 0. Now by the definition of ≥ on Γ , we have v(x)≥ 0= v(1) just if x = x1−1 ∈ A, soA′ =A.

23 In multiplicative notation, this would be called a “group with zero,” with the absorbing element∞ playing the role of “zero.”

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Suppose on the other hand v : K× Γ is a valuation, with valuation ring A = 0 ∪ x ∈ K× : v(x) ≥ 0 andvalue group Γ . Writing U = ker(v), and π : K× K×/U for the natural map, there is a canonical isomorphismφ : K×/U

∼−→ Γ such that v = φ π. The field of fractions of A is K , so [5.30] gives a valuation v ′ : K× K×/A×.Now A× =

x ∈ A\0 : x−1 ∈ A

= x ∈ K× : v(x) ≥ 0 & − v(x) = v(x−1) ≥ 0 = x ∈ K× : v(x) = 0 = U ,so v ′ = π. Thus v = φ π = φ v ′, so v ′ is canonically equivalent to v. Finally, v(x) ≤ v(y) ⇐⇒ 0 = v(1) =v(x x−1) = v(x)− v(x)≤ v(y)− v(x) = v(y x−1) ⇐⇒ y x−1 ∈ A\0 ⇐⇒ v ′(x)≤ v ′(y) by the definition of v ′ in[5.30], so the order is preserved.

32. Let Γ be a totally ordered abelian group. A subgroup∆ of Γ is isolated in Γ if, whenever 0≤β≤ α and α ∈∆, we haveβ ∈∆. Let A be a valuation ring of a field K, with value group Γ (Exercise 31). If p is a prime ideal of A, show that v(A\p)is the set of elements≥ 0 of an isolated subgroup∆ of Γ , and that the mapping so defined of Spec(A) into the set of isolatedsubgroups of Γ is bijective.

Write∆+ = v(A\p). Obviously 1 /∈ p, so v(1) = 0 ∈∆+. If α= v(a) andβ= v(b ) are in∆+, with a, b /∈ p, thensince p is prime ab /∈ p, and so α+β= v(a)+v(b ) = v(ab ) ∈∆+. Thus∆+ is a submonoid of Γ and∆=∆+ ∪−∆+is a subgroup whose elements ≥ 0 are∆+

Suppose 0 ≤ β ≤ α in Γ with α ∈ ∆+. If β = 0 or α, then β ∈ ∆+, so assume not. Let α = v(a) for a ∈ A\pand α−β = v(c) for c ∈ A. If c /∈ p, then α−β ∈∆+, so β−α ∈∆ and β ∈∆+, since ∆ is a subgroup. If c ∈ p,consider b = ac−1. Now v(b ) = α− (α−β) =β> 0, so b ∈ A, but b /∈ p, since otherwise b c = a ∈ p, contrary toassumption. Thus β ∈∆+.

The correspondence is injective, for assume p, q in Spec(A) are such that ∆(p) =∆(q). Then for every x ∈ A\pthere is y ∈A\qwith v(x) = v(y). Then 0= v(x)−v(y) = v(xy−1), so xy−1 ∈A× and x = (xy−1)y ∈A×(A\q) =A\q,so A\p⊆A\q. Symmetrically, A\q⊆A\p, so p= q.

For surjectivity, let an isolated subgroup ∆ be given. The natural candidate for ∆ = v(A\p) is p = A\v−1(∆).Certainly it has the right image. Since∞ /∈∆ we get 0 ∈ p. If x ∈ p, then v(−x) = v(x) /∈∆, so −x ∈ p. If x, y ∈ p,then v(x+y)≥min

v(x), v(y)

>∆, so x+y ∈ p. Finally, if x, y /∈ p, then v(x), v(y) ∈∆, so v(xy) = v(x)+v(y) ∈∆, and xy /∈ p.

If p is a prime ideal of A, what are the value groups of the valuation rings A/p, Ap?

For A/p, define v : A/p→ Γ ∪∞ by v(x) =¨

v(x), x /∈ p,∞, x ∈ p.

Assuming it is well defined, it inherits axioms (1)

and (2) of [5.31] from v. To see it is well defined, assume x − y ∈ p. Then

v(x) =min

v(x),∞

=min

v(x), v(y − x)

≤ v

x +(y − x)

= v(y),

and similarly v(y)≤ v(x). Then we can extend v to the fraction field k of A/p, and it gives a valuation v : k×→∆.Since units of A/p are images of units of A (since in the quotient m 7→ m/p are the maximal ideals) and only theseare taken to 0 by v, we see A/p is the valuation ring of v, and∆ is the valuation group of A/p.

For Ap, the group K× is unchanged. Since A is local, p⊆m, meaning U =A\m⊆A\p⊆Ap\pAp =: Up, the unitsof Ap. Thus the value group is a further quotient of Γ =K×/U . We can write an element of Up as a/b for a, b ∈A\p,so∆= v(A\p) = v(Up). By the third isomorphism theorem (2.1.i), K×/Up

∼= (K×/U )/(Up/U ) = Γ/v(Up) = Γ/∆.

33. Let Γ be a totally ordered abelian group. We shall show how to construct a field K and a valuation v of K with Γ as valuegroup. Let k be any field and let A= k[Γ ] be the group algebra of Γ over k. By definition, A is freely generated as a k-vectorspace by elements xα (α ∈ Γ ) such that xαxβ = xα+β. Show that A is an integral domain.

This will follow from our proof of (1) in the next paragraph.

If u = λ1xα1+ · · ·+λn xαn

is any non-zero element of A, where the λi are all 6= 0 and α1 < · · ·< αn , define v0(u) tobe α1. Show that the mapping v0 : A\0→ Γ satisfies conditions (1) and (2) of Exercise 31.

(1) Let f =∑

aαxα and g =∑

bβxβ be non-zero elements of A, with v0( f ) = α0 and v0(g ) = β0. Then in f g ,the non-zero coefficient of lowest index is that of xα0+β0

, which is aα0bβ0

, since for all other pairs α, β of indices wehave α+β>α0+β0. As k is a field, aα0

bβ06= 0, so f g 6= 0 and v( f g ) = α0+β0.

(2) Let f =∑

aαxα and g =∑

bβxβ be nonzero elements of A, with v0( f ) = α0 and v0(g ) = β0; then thelowest potentially nonzero coefficient index of f + g is α0 +β0. (Of course, there could be cancellation.) Thusv0( f + g )≥min

v0( f ), v0(g )

.

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Let K be the field of fractions of A. Show that v0 can be uniquely extended to a valuation v of K, and that the valuegroup of v is precisely Γ .

Axiom (1) requires that 0= v( f / f ) = v0( f )+ v(1/ f ), so that v( f −1) =−v0( f ) for all nonzero f ∈A. Then forf /g ∈K with f , g ∈A we must have v( f /g ) = v0( f )− v0(g ), so the extension v is unique, if the definition definesa valuation. Suppose f /g = f ′/g ′ in K , for f , f ′, g , g ′ ∈ A. Then by the definition of localization, f g ′ = f ′ g , sov0( f ) + v0(g

′) = v0( f′) + v0(g ), and v( f /g ) = v0( f )− v0(g ) = v0( f

′)− v0(g′) = v( f ′/g ′), so v is well defined.

Evidently v(K) = v(A)− v(A) = Γ − Γ = Γ . It remains to verify axiom (2). Again let f /g , f ′/g ′ ∈ K be given, forsome f , f ′, g , g ′ ∈A. Then h := f

g +f ′

g ′ =f g ′+ f ′ g

g g ′ , and

v(h) = v( f g ′+ f ′ g )− v(g g ′)≥min

v( f )+ v(g ′), v( f ′)+ v(g )

−

v(g )+ v(g ′)

=min

v( f )− v(g ), v( f ′)− v(g ′)

=min

v( f /g ), v( f ′/g ′)

.

It should be pointed out that v(A) = v

k(Γ )

= Γ already, so A is not the valuation ring of K . Rather, B =0 ∪ x ∈K : v(x)≥ 0 is, by [5.31].

34. Let A be a valuation ring and K its field of fractions. Let f : A→ B be a ring homomorphism such that f ∗ : Spec(B)→Spec(A) is a closed mapping. Then if g : B → K is any A-algebra homomorphism (i.e., if g f is the embedding of A inK) we have g (B) =A.

Since A = g

f (A)

⊆ K we have A ⊆ g (B) =: C . Then C is a valuation ring, by (5.18.ii). Let o be a maximalideal of C ; since g |C : B C is surjective, g ∗(o) = n is a maximal ideal of B . By [1.18.i], n ⊆ Spec(B) is closed;as f ∗ is a closed mapping, f ∗(n) is closed, so by [1.18.i] again it must be a singleton containing a maximal ideal.As A is local, that ideal is the unique maximal ideal m Ã A. Since g f : A ,→ K is the inclusion, this means thatm= f ∗(n) = f ∗

g ∗(o)

= (g f )∗(o) = o∩A, so o dominates m. By [5.27] (valuation rings are domination-maximal),this shows A=C and m= o.

35. From Exercises 1 and 3 it follows that, if f : A→ B is integral and C is any A-algebra, then the mapping ( f ⊗1)∗ : Spec(B⊗AC )→ Spec(C ) is a closed map.

Conversely, suppose that f : A→ B has this property and that B is an integral domain. Then f is integral.Write A′ = f (A) ⊆ B , and K for the field of fractions of B . To show B is integral over A′, it is enough to show

B is in the integral closure of A′ in K . By (5.22) it is enough to show B is in each valuation ring of K containingA′. Let C be one such. Then A′ ⊆ B , C ⊆ K . Multiplication B ×C → K is A-bilinear (equivalently, A′-bilinear), soinduces a map g : B ⊗AC →K . Now f ⊗ idC : A⊗AC → B ⊗AC , and (2.14) gives an isomorphism φ : C

∼−→A⊗AC .Write F = ( f ⊗ idC )φ, so that F ∗ : Spec(B⊗AC )→ Spec(C ) is closed, by assumption. The composition g F takesc 7→ 1A⊗ c 7→ 1B ⊗ c 7→ c , and so is the inclusion C ,→ K . The preceding [5.34] then says that g (B ⊗AC ) = C . Inparticular, for each b ∈ B we have b = g (b ⊗ 1C ) ∈C , so B ⊆C .

Show that the result just proved remains valid if B is a ring with only finitely many minimal prime ideals (e.g., if Bis Noetherian).

First, if B is Noetherian, (7.13) says the (0) ideal has a primary decomposition, and (4.6) says the finitely manyisolated primes of this decomposition are precisely the minimal ideals of B .

Second, the statement needs some clarification. The hypothesis being replaced (by B only having finitely manyminimal prime ideals) is that B is integral, not the closed mapping assumption.

Now suppose f : A→ B satisfies the assumption. and B has only finitely many minimal prime ideals p1, . . . , pn .The surjections πi : B B/pi give rise to compositions πi f : A → B B/pi . Let an A-algebra C be given.Since tensor is left exact and B B/pi is a surjection, gi : B ⊗AC → (B/pi ) ⊗AC is a surjection. By [1.21.iv],g ∗i : Spec

(B/pi )⊗AC

→ Spec(B ⊗AC ) is a closed map, and by assumption, Spec(B ⊗AC )→ Spec(C ) is a closedmap, so composing, Spec

(B/pi )⊗AC

→ Spec(C ) is closed. As C was arbitrary, πi f has the property above, andsince B/pi is an integral domain,πi f is integral. By [5.6], the map (π1 f , . . . , πn f ) : A→

∏

B/pi is integral; thismap factors as (π1, . . . , πn) f : A→ B→

∏

B/pi . The kernel of the homomorphism (π1, . . . , πn) : B→∏

B/pi isthe nilradical N=

⋂

pi of B , so we have a factorization A→ B B/N∏

B/pi . Since∏

B/pi is integral over theimage of A, so is the embedded subring B/N. Now let x ∈ B ; then its image x ∈ B/N satisfies a monic polynomialequation x m +

∑

j<m b j x j = 0 for some b j ∈ f (A); lifting, this means p(x) = x m +∑

j<m b j x j ∈N. Then there isan integer l large enough that p(x)l = 0 in B ; but p(x)l is a monic polynomial in f (A)[x], so x is integral over f (A).

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Chapter 6: Chain Conditions

Jordan–Hölder Theorem. Consider an A-module M of finite length. (6.7) says that every composition series of Mhas the same length, and the book claims (p. 77) that the multiset of isomorphism classes of quotients of successiveterms is the same for any choice of composition series. The proof, it goes on, is the same as for finite groups. Werecall it here.1

The proof proceeds by induction on the length l (M ) of M . If l (M ) = 0 or 1, we are done. Assume inductivelythat the result holds for all modules of length n, and let l (M ) = n+ 1. Assume M has the two composition series

M =M0 )M1 ) · · ·)Mn+1 = 0, M =N0 )N1 ) · · ·)Nn+1 = 0.

If M1 = N1, then by the inductive hypotheses the multisets S = Mi/Mi+1ni=1 and T = Ni/Ni+1ni=1 of quotientsare equal so since M/M1 =M/N1, the quotient multisets of the two composition series for M are equal.

If M1 6=N1, let P1 = M1 ∩N1. Note that M1 ( M1+N1 ⊆ M , so since M/M1 was assumed simple, M1+N1 = M .Now M1/P1 =M1/(M1∩N1)∼= (M1+N1)/N1 =M/N1 by the second isomorphism theorem (2.1.ii), and this quotientis simple. Symmetrically, N1/P1

∼=M/M1. By the proof of (6.7), l (P1)≤ l (M1) = n is finite, so P1 has a compositionseries P1 ) P2 ) · · · ) Pp = 0. Write U for the quotient multiset. M1 ) P1 ) · · · ) Pp = 0 is a compositionseries for M1. Since l (M1) = n, we have p = n, and by the induction hypothesis, the multiset M1/P1 ∪ U =M/N1 ∪ U is the same as the multiset S = Mi/Mi+1ni=1. Then the quotient multiset for the Mi compositionseries of M is M/M1 ∪ S = M/M1, M/N1 ∪U . Similarly N1 ) Pi is a composition series for N1 with quotientmultiset N1/P1 ∪U = M/M1 ∪U , by inductive assumption equal to the multiset T = Ni/Ni+1ni=1. Then theNi composition series for M yields the quotient multiset M/N1 ∪T = M/N1, M/M1 ∪U as well.

EXERCISES1. i) Let M be a Noetherian A-module and u : M →M a module homomorphism. If u is surjective, then u is an isomorphism.

For all n ≥ 0, we have ker(un) a submodule of M and ker(un) ⊆ ker(un+1); as M is Noetherian, we eventuallyhave ker(un) = ker(un+1). Any element y ∈ im(un) is un(x) for some x ∈M , and if 0= u(y) = u

un(x)

= un+1(x),then x ∈ ker(un+1) = ker(un), so y = un(x) = 0 already. Thus u is injective on im(un). But since u is surjective,im(un) =M , so u is injective, hence an isomorphism.

ii) If M is Artinian and u is injective, then again u is an isomorphism.For all n ≥ 0, we have im(un) a submodule of M and im(un) ⊇ im(un+1); as M is Artinian, we eventually

have im(un) = im(un+1). For each element x ∈ M we have un(x) ∈ im(un) = im(un+1), so there is y ∈ M withun(x) = un+1(y) = un

u(y)

. As u is injective, un is also injective, so x = u(y). As x was arbitrary, u is surjective,hence an isomorphism.

2. Let M be an A-module. If every non-empty set of finitely generated submodules of M has a maximal element, then M isNoetherian.

By (6.2), it will suffice to show any submodule N of M is finitely generated. Let Σ be the set of finitely generatedsubmodules of N . By assumption,Σ has a maximal element N0. If N0 (N , there is x ∈N\N0, and then N0+Ax )N0is a finitely generated submodule of N , contradicting maximality of N0. Thus N0 =N is finitely generated.

1 http://planetmath.org/encyclopedia/ProofOfTheJordanHolderDecompositionTheorem.html

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Chapter 6: Chain Conditions Ex. 6.3

3. Let M be an A-module and let N1, N2 be submodules of M . If M/N1 and M/N2 are Noetherian, so is M/(N1 ∩N2).Similarly with Artinian in place of Noetherian.

By the second isomorphism theorem (2.1.ii), N1/(N1 ∩N2)∼= (N1+N2)/N2. Since (N1+N2)/N2 is a submoduleof the Noetherian (resp. Artinian) M/N2, (6.3.i) (resp. (6.3.ii)) shows N1/(N1 ∩N2) is Noetherian (resp. Artinian).Now the third isomorphism theorem (2.1.i) gives an exact sequence 0→N1/(N1∩N2)→M/(N1∩N2)→M/N1→ 0.The outside terms are Noetherian (resp. Artinian), so another use of (6.3.i) (resp. (6.3.ii)) shows that M/(N1∩N2) isNoetherian (resp. Artinian).

4. Let M be a Noetherian A-module and let a be the annihilator of M in A. Prove that A/a is a Noetherian ring.M is finitely generated by (6.2), say by x1, . . . , xn . Then if ai =Ann(xi ), we have A/ai

∼=Axi ; as a submodule ofa Noetherian module, it is, by (6.3.i), also Noetherian. Since a=

⋂

ai , by [6.3] and induction, A/a is a NoetherianA-module, hence a Noetherian A/a-module.

If we replace “Noetherian” by “Artinian” in this result, is it still true?No. Let p ∈N be a non-zero prime and consider Example 3) of p. 74, the p-quasicyclic group G =Z/p∞Z.2 It is

an abelian group, so aZ-module, and the book states that it is Artinian.3 It is generated by the elements xn =Z+1/pn ,and AnnZ(xn) = (p

n), so AnnZ(G) = (0). Now Z∼=Z/(0) is not an Artinian ring by Example 2) of p. 74.Our proof for the Noetherian case above fails precisely because there are infinitely many xn ; we do have each

Z/(pn) =Z/AnnZ(xn)Artinian, being isomorphic to the submodule Gn of G, but the result of [6.3] does not extendto infinite intersections.

5. A topological space X is said to be Noetherian if the open subsets of X satisfy the ascending chain condition (or, equivalently,the maximal condition). Since closed subsets are complements of open subsets, it comes to the same thing to say that theclosed subsets of X satisfy the descending chain condition (or, equivalently, the minimal condition). Show that, if X isNoetherian, then every subspace of X is Noetherian, and that X is compact.

Let Y ⊆ X be a subspace; in the subspace topology, the open sets of Y are precisely the intersections with Y ofopen sets of X. If ⟨Vn⟩ is an ascending chain of open subsets of Y , let Un ⊆ X be open and such that Vn = Un ∩Y .Then there is some n such that Un = Un+1 = · · · , so intersecting with Y , we get Vn =Vn+1 = · · · . This shows Y isNoetherian.

LetU be an open cover of X, and let Σ be the collection of all finite unions of elements ofU . By the maximalcondition on opens, Σ has a maximal element V . If there is x ∈ X \V , there is some U ∈U containing x sinceUis an open cover, and then U ∪V ∈ Σ strictly contains V , contradicting maximality. Thus X =V is a finite unionof elements ofU . This shows X is compact.

6. Prove that the following are equivalent:i) X is Noetherian.ii) Every open subspace of X is compact.iii) Every subspace of X is compact.

i) =⇒ iii): This follows from [6.5]: each Y ⊆X is itself Noetherian, and each Noetherian space is compact.iii) =⇒ ii): This is trivial: each open subspace is a subspace.ii) =⇒ i): Let U1 ⊆U2 ⊆ · · · be an ascending chain of open subsets of X, and U =

⋃

n∈NUn . Since U is compact,U is a union of a finite set Un1

, . . . , Unm. But then if n =max j n j , we see U =Un .

2 See http://planetmath.org/encyclopedia/QuasicyclicGroup.html. G is the group of elements of Q/Z with denominator a p-power, or equivalently Z[1/p]/Z. Taking its image under x 7→ e2πi x gives an isomorphism to the subgroup z : ∃n ≥ 0 (z pn

= 1) of C×. Thusit is an increasing union of the groups of (pn)th roots of unity, or equivalently the direct limit of the system (Z/pnZ, πmn), where for m ≤ n wehave πmn : Z/p mZ→Z/pnZ taking 1 7→ pn−m .

3 are its only proper subgroups. First, we claim that if x ∈ Gn \Gn−1, then ⟨x⟩ = Gn . Since x = Z+ a/pn for some a ∈ Z\(p), we have(a)+(p) = (1) in Z. By (1.16), (a)+(pn) = 1, so there are b , m ∈Z such that ba+m pn = 1. Thus (ba/pn) = m+(1/pn) in Z[1/p], so b x = xnand ⟨x⟩ = Gn . Now suppose H is a proper subgroup of G. Since G =

⋃

Gn , we see H fails to contain some Gn+1. Let n + 1 be minimal suchthat this happens. Then H contains no element of Gn+1\Gn by the work above, but by assumption contains all of Gn , so H = Gn . Now G isArtinian, for given a strictly descending chain of submodules starting with G, the second module is some Gn , and Gn properly contains onlyn− 1 submodules.

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Ex. 6.7 Chapter 6: Chain Conditions

7. A Noetherian space is a finite union of irreducible closed subspaces. Hence the set of irreducible components of a Noetherianspace is finite.

Recall from [1.19] that a topological space C is irreducible if for every pair of nonempty open subsets U1, U2,we have U1 ∩U2 6= ∅. Taking complements Fi = C \Ui , this means for every pair of closed subsets F1, F2 ( C , wehave C 6=C \(U1 ∩U2) = (C \U1)∪ (C \U2) = F1 ∪ F2. That is, C is not a union of proper closed subspaces.

Suppose, for a contradiction, that the result is false. Then there is a Noetherian space X such that X is an elementof the set Σ of closed subsets of X that are not unions of finitely many irreducible closed subspaces. Since Σ isnonempty and X is Noetherian,Σ has a minimal element C . Since C is not a finite union of irreducible sets, it is notitself an irreducible set. Thus it is reducible, and so a union of two proper closed subspaces F1 and F2. But F1 and F2are both finite unions of irreducible closed sets, so C is as well, a contradiction.

Recall from [1.20.iii] that the irreducible components of a space X are the maximal irreducible subsets of X, andthat they are closed and cover X. Since a Noetherian space X is a union of finitely many irreducible closed subspaces,it is a fortiori a union of finitely many maximal such, so it is a union of finitely many irreducible components. Let nbe the minimal possible number needed to cover X, and let C1, . . . , Cn be irreducible components covering X. If F isany other irreducible closed set, then F =

⋃nj=1(F ∩C j ) expresses F as a union of closed subsets; as F is irreducible,

F ⊆C j for some j . Thus C1, . . . , Cn are the only irreducible components of X.

8. If A is a Noetherian ring, then Spec(A) is a Noetherian topological space. Is the converse true?Every closed subset of Spec(A) is ([1.15]) of the form V (a) for some radical ideal a Ã A. Let

V (a j )

j∈N bean infinite descending chain collection of closed subsets of Spec(A). Since V (a j+1) ⊆ V (a j ), taking intersectionsof these sets of primes (recalling the a j are radical; see (1.14)) gives a j ⊆ a j+1. Since A is Noetherian, eventuallyan = an+1 = · · · , and so V (an) =V (an+1) = · · · ; thus Spec(A) is Noetherian.

The converse is not true. Let k be a field, and A = k[x1, x2, . . .] a polynomial ring over k in countably manyindeterminates. Let any sequence (mn)

∞n=1 of integers> 1 be given. Let b be the ideal generated by x m1

1 and xn−x mn+1n+1

for all n ≥ 1. Write yn = xn in B = A/b. Then y m11 = 0 and yn = y mn+1

n+1 for all n ≥ 1. If p = (y1, y2, . . .), we haveB/p∼= k, so p is maximal. Now yn+1 ∈ r (yn) and y1 ∈ r (0), so p⊆N(B)⊆ p, showing p is the unique minimal primeas well. Since all primes then contain p, which is maximal, p is the only prime of B . Thus Spec(B) = p is obviouslyNoetherian. But (y1)( (y2)( (y3)( · · · is an infinite ascending chain of ideals, so B is not Noetherian.

Relatedly4, but using material from earlier in the book, let k be a field and Γ a non-zero totally ordered groupwith only finitely many isolated subgroups and such that Γ>0 := γ ∈ Γ : γ > 0 has no least element (for exampletake Z[1/p] ⊆ Γ ⊆ R for p ≥ 2). Let K be the field of fractions of the group algebra k[Γ ]. Then [5.33] gives asurjective valuation v : K×→ Γ , and A= 0 ∪ x ∈ K : v(x)≥ 0 is the associated valuation ring. [5.32] shows thatA has only finitely many prime ideals. For any γ ∈ Γ+, there exists x ∈ k[Γ ] with v(x) = γ . Elements y ∈ (x) havevalue v(y)≥ v(x) by axiom (1) of [5.31], so if x, y ∈A have v(x)< v(y), it is impossible that x ∈ (y), and by [5.28]we have (y) ( (x). Now any infinite decreasing sequence γ1 > γ2 > · · · > 0 in Γ gives rise to an infinite increasingsequence of ideals of A.

Simpler5, let k be a field, A = k[x1, x2, . . .] a polynomial ring over k in countably many indeterminates, c =(x2

1 , x22 , · · · ), and C =A/c. Write z j = x j and q= (z1, z2, . . .). Then C/q∼= k, so q is maximal, and q⊆N(C )⊆ q, so

q is minimal, and thus Spec(C ) = q. But (z1)( (z1, z2)( (z1, z2, z3)( · · · is an infinite ascending chain of ideals.

9. Deduce from Exercise 8 that the set of minimal prime ideals in a Noetherian ring is finite.Let A be a Noetherian ring. By [6.8], X is a Noetherian space. By [6.7], X has only finitely many irreducible

components. By [1.20.iv], the irreducible components of X = Spec(A) are the closed sets V (p) for p a minimalprime; thus there are only finitely many minimal primes of A.

10. If M is a Noetherian module (over an arbitrary ring A) then Supp(M ) is a closed Noetherian subspace of Spec(A).Recall ([3.19]) that Supp(M ) is the set of prime ideals p Ã A such that Mp 6= 0. Write a = Ann(M ). By [3.19.v],

Supp(M ) =V (a) is closed. [1.21.iv] gives a homeomorphism V (a)≈ Spec(A/a). But by [6.4], A/a is a Noetherianring, so [6.8] shows Spec(A/a)≈V (a) is a Noetherian space.

4 http://pitt.edu/~yimuyin/research/AandM/exercises06.pdf5 [KarpukSol]

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Chapter 6: Chain Conditions Ex. 6.11

11. Let f : A→ B be a ring homomorphism and suppose that Spec(B) is a Noetherian space (Exercise 5). Prove that f ∗ : Spec(B)→Spec(A) is a closed mapping if and only if f has the going-up property (Chapter 5, Exercise 10).

In [5.10.i]we showed that f ∗ being closed implies f has the going-up property. Now suppose Spec(B) is Noethe-rian and f has the going-up property. Let V (b), for bÃ B radical, be an arbitrary closed set of Spec(B). Then by [6.5],V (b) is itself Noetherian, By [1.21.iv], V (b)≈ Spec(B/b), so by [6.7], Spec(B/b) has only finitely many irreduciblecomponents. By [1.20.iv], these correspond to minimal primes of B/b, and hence minimal elements q j of V (b). Letp j = qc

j . Now (1.18) shows a := bc =⋂

q j

c =⋂

qcj =

⋂

p j . Now if p ∈ V (a) we have⋂

p j ⊆ p, so by (1.10.ii),p j ⊆ p for some j . Thus V (a) =

⋃

V (p j ). [5.10.i] shows that f ∗

V (q j )

= V (p j ), so f ∗

V (b)

= f ∗⋃

V (q j )

=⋃

f ∗

V (q j )

=⋃

V (p j ) =V (a) is closed.

12. Let A be a ring such that Spec(A) is a Noetherian space. Show that the set of prime ideals of A satisfies the ascending chaincondition. Is the converse true?

Let p1 ⊆ p2 ⊆ · · · be an ascending chain of prime ideals of A. Then V (p1)⊇V (p2)⊇ · · · is a descending chain ofclosed subsets of Spec(A). Since Spec(A) is Noetherian, the descending chain terminates in some V (pn) =V (pn+1).Then pn ∈V (pn) =V (pn+1), so pn+1 ⊆ pn ⊆ pn+1 and the chain stabilizes.

The converse is untrue. Let X be an infinite set; its power setP (X )with the partial order given by⊆ is a Booleanalgebra by the proof of [1.25]. Let A be the associated Boolean ring ([1.24]). (Note that in fact A∼=

∏

X F2.) If a ∈A,then the principal ideal (a) Ã A is the set of ba = b ∩ a for b ∈ P (X ), which is the set of subsets b ⊆ a. Letx1, x2, . . . be an infinite sequence of distinct elements of X, and for each n let pn be the principal ideal generated bythe element X \xn of A. Each pn is prime, for if a, b ∈A\pn , then xn ∈ a and xn ∈ b , so xn ∈ a∩ b = ab , meaningab /∈ pn . If we let Sn = s1, . . . , sn for each n ≥ 1, then Sn ( Sn+1 for each n, so if an is the principal ideal (Sn), thenan ( an+1 for each n, so any prime containing the latter contains the former, and V (an)⊇V (an+1) in Spec(A). Butxn+1 ∈ Sn+1 ∈ an+1, so an+1 6⊆ pn+1, while for a ∈ an we have xn+1 /∈ Sn ⊇ a, so that an ⊆ pn+1. Thus we have astrict containment V (an))V (an+1) for each n, so the V (an) are an infinite descending sequence of closed subsets ofSpec(A), which is then not Noetherian. But by [1.11.ii], every prime of A is maximal, so each chain of prime idealsof A has length zero, and the set of prime ideals of A satisfies the ascending chain condition.

For another demonstration the converse is untrue,6 let k be a field, B =∏∞

j=1 k the product of countably manycopies of k, and A= k · 1+

⊕

k the subring of B consisting of all eventually constant sequences of elements of k.Write e j ∈ A for the element with a 1 at the j th place and 0 elsewhere, and fn = 1−

∑

j≤n e j . For each n ≥ 0 thesubring An =

∑

j<n ke j + k fn ( A is the set of sequences of elements of k constant from the (n+ 1)th element on,and we have a natural isomorphism Bn

∼−→ kn+1 taking e j 7→ e j for j ≤ n and fn 7→ en+1. If b j =∑

i 6= j (ei ), the proofof [1.22], shows that Spec(kn+1) is a disjoint union of sets of ideals b j + pe j for p a prime of k; since k is a field,Spec(kn+1) = b1, . . . , bn+1. For the corresponding primes of Bn write pn, j =

∑

i 6= j (ei ) + ( fn) for j = 1, . . . , n andqn =

∑

j≤n(e j ). Let p ∈ Spec(B). Then p∩An is a prime. One possibility is that this prime is qn for each n. SinceB =

⋃

Bn , then p0 := p =⋃

qn =⊕

k, and B/p0 = B/⊕

k ∼= k, so p0 is maximal.7 Otherwise there is some Anwith p ∩An = pn, m , and it follows p ∩Aj = p j , m for all j ≥ n. Thus e j ∈ p for j > m and fm ∈ p, so p containsp j =

∑

n 6= j (e j )+( f j ). As p j is the kernel of the projection of A onto the j th coordinate, it follows again p j is maximal,so p= p j . All primes of A being maximal, It follows any ascending sequence of prime ideals of A is constant. On theother hand qn =

∑

j≤n(e j ) gives an infinite ascending sequence of ideals of A. Evidently V (qn)⊇V (qn+1), and sinceqn ⊆ pn+1 but qn+1 6⊆ pn+1, the inclusion is strict. Thus Spec(A) is not Noetherian.

6 http://pitt.edu/~yimuyin/research/AandM/exercises06.pdf7 In case this wasn’t clear, since the diagram

0 // qn // _

An _

// // k // 0

0 // qn+1 // An+1

// // k // 0is commutative, [2.18] and [2.19] give a short exact sequence 0→ p0 ,→A k→ 0 of direct limits.

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Chapter 7: Noetherian Rings

Theorem 7.5*. If A is Noetherian, then the formal power series ring A[[x]] is Noetherian.If f = ax m+(deg> m) is an element of A[[x]], define ord( f ) = m. Let a be an ideal of A[[x]], and let l be the set of

trailing coefficients of series in a. This is an ideal of A, for if a, b ∈ l and c ∈A, there are elements f = ax m+(deg> m)and g = bx p+(deg> p) of a; without loss of generality, assume p ≥ m; then the trailing coefficient of x p−m f −g ∈ ais a− b , and the trailing coefficient of c f ∈ a is ca. Since A is Noetherian, l is finitely generated, say by a1, . . . , an . Bythe definition of l, for i = 1, . . . , n there is a series fi ∈A[[x]] of the form fi = ai x ri +(deg> ri ). Let r =maxn

i=1 ri .The fi generate an ideal a′ ⊆ a in A[[x]].

Let f be an arbitrary element of a, with m = ord( f ) ≥ r . If gi , 0 = 0 for all i , we have f = f −∑

gi , 0 fi oforder m. Let p ≥ m, and assume inductively that there are polynomials gi , p ∈ A[x] with deg(gi , p ) ≤ m − ri

such that f ′ = f −∑

gi fi has ord( f ′) = p. If f ′ = ax p + (deg > p) write a =∑n

i=1 ui ai , where ui ∈ A; thenf ′ −

∑

(ui x p−ri ) fi = f −∑

(gi , p + ui x p−ri ) fi is in a and has degree > p. Then gi , p+1 = gi , p + ui x p−ri has degree≤ p − ri . Since any terms we might add in transforming gi , p to gi , p+1 are of strictly higher degree, as p →∞, thepolynomials gi , p ∈A[x] converge to some gi ∈A[[x]], and f −

∑

gi fi = 0, so f ∈ a′.If we write M =A+Ax+ · · ·+Ax r−1, then this shows that for any f ∈ a, there is g ∈ a′ such that f − g ∈ a∩M .

Now M is a finitely generated A-module, so it is Noetherian by (6.5), and thus a∩M is a finitely generated A-moduleby (6.2). Therefore a= (a∩M )+ a′ is finitely generated.

EXERCISES1. Let A be a non-Noetherian ring and let Σ be the set of ideals in A which are not finitely generated. Show that Σ has

maximal elements and that the maximal elements of Σ are prime ideals.Hence a ring in which every prime ideal is finitely generated is Noetherian (I. S. Cohen).This result has actually be vastly generalized: there is a metatheorem giving results of the form “any ideal maximal

with respect to not having property P is prime” for large natural classes of properties P of ideals.1 For example, forany infinite cardinal κ, any ideal maximal with respect to not being generated by < κ elements is prime (it is nothowever guaranteed that such ideals exist); the result we prove here is the κ=ℵ0 case.

Since A is not Noetherian, Σ is not empty. Let ⟨aα⟩α be a chain in Σ, and a its union. If a was finitely generated,say by xi ∈ aαi

(1 ≤ i ≤ n) with α1 ≤ · · · ≤ αn , then we would have a = (x1, . . . , xn) ⊆ aαn⊆ a, showing aαn

/∈ Σ, acontradiction. Then Zorn’s Lemma furnishes maximal elements of Σ.

It is not harder to prove the next part for ideals generated by <κ elements than <ℵ0 elements, so redefine Σ tobe the set of ideals not generated by < κ elements. Let p be a maximal element of Σ; by the last paragraph, Σ hasmaximal elements if κ = ℵ0. Suppose a /∈ p and b ∈ A are such that ab ∈ p; we show b ∈ p. Now (a) + p ∈ Σ. Ifit is generated by the elements (bα + xα) for bα ∈ A and xα ∈ p (α < κ), then (a) + p = (a) + (xα). Write a = (xα).Now if y ∈ p\a, then y ∈ p∩ (a), so there is z ∈ A, such that az = y ∈ p. It follows that z ∈ (p : a), so y ∈ a(p : a).Thus a+ a(p : a) = p). If (p : a) /∈Σ, we would have a(p : a) /∈Σ and hence p /∈Σ, so it follows that (p : a) ∈Σ. Nowp⊆ (p : a), and if the containment were strict, we would have (p : a) /∈Σ, so b ∈ (p : a) = p.

2. Let A be a Noetherian ring and let f =∑∞

n=0 an xn ∈A[[x]]. Prove that f is nilpotent if and only if each an is nilpotent.By [1.5.ii], if f is nilpotent, then all an are nilpotent. On the other hand, suppose all an are nilpotent. By (7.15), the

nilradical N of A is nilpotent, meaning there is m ≥ 1 such that Nm = 0. That means any product aI =∏m

i=1 ani= 0.

In f m , each term is divisible by some aI , so f m = 0.

1 Tsit Yuen Lam and Manuel L. Reyes: http://bowdoin.edu/~reyes/oka1.pdf

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Chapter 7: Noetherian Rings Ex. 7.3

3. Let a be an irreducible ideal in a ring A. Then the following are equivalent:i) a is primary;ii) for every multiplicatively closed subset S of A we have (S−1a)c = (a : x) for some x ∈ S;iii) the sequence (a : xn) is stationary, for every x ∈A.

i) =⇒ ii): Let a be primary and a ∈ a. (4.8) says that either (S−1a)c = a= (a : 1) or (S−1A)c =A= (a : a).ii) =⇒ iii): Let x ∈A, and Sx = 1, x, x2, . . .. Recall from [4.12] that Sx (a) := (S

−1x a)c = y ∈A : Sx y ∩ a 6=∅.

But this is⋃

s∈Sx(a : s) =

⋃

n≥0(a : xn). By assumption, this is also equal to (a : s) for some s = xn ∈ S. Then(a : xn)⊆ (a : xn+m)⊆ Sx (a) = (a : xn) for all m ≥ 0, so the chain of ideals (a : xn) is stationary.

iii) =⇒ i): Since a is irreducible in A, we have (0) irreducible in A′ = A/a. Suppose exey ∈ a with ex, ey ∈ Aand y /∈ a. Let x, y be their images in A′. The increasing chain (a : exn) in A is stationary, and its image in A′ is(0 : xn) = Ann(xn), so for some n we have Ann(xn) = Ann(xn+1). If a ∈ (y) ∩ (xn), write a = b y = c xn . Then0= b xy = c xn+1, so c ∈Ann(xn+1) =Ann(xn), meaning a = c xn = 0. Thus (0) = (y)∩(xn). Since (0) is irreducible,it follows that (xn) = (0), so exn ∈ a. Thus a is primary.

4. Which of the following rings are Noetherian? In all cases the coefficients are complex numbers.i) The ring of rational functions of z having no pole on the circle |z |= 1.

This ring A is Noetherian. Any element of A can be written in least terms as p(z)/q(z) ∈C(z), for p(z), q(z) ∈C[z] such that q(z) has no root on the circle |z | = 1. Put another way, q(z) can be any polynomial in the multi-plicatively closed set S = C[z]\

⋃

|a|=1(z − a). Thus A= S−1C[z]. Since C[z] is Noetherian (for example by (7.5)),its localization A is Noetherian by (7.3).

ii) The ring of power series in z with a positive radius of convergence.This ring A is Noetherian. Let B be an arbitrary ring, and z an indeterminate. Recall ([1.5]) that f ∈ B[[z]] is

a unit just if f ∈ B× + (z), that is, the constant term of f is a unit. If B = k is a field, this just means the constantterm is nonzero. Each nonzero element f ∈ k[[z]] has some order ord( f ) (the least m such that the coefficient ofz m in f is non-zero), and thus f = zord( f ) g for some g ∈ k×+ (z). Since g is a unit, it follows f g−1 = zord( f ) ∈ ( f ),and so ( f ) = (zord( f )). Thus the ideals of k[[z]] are (0) and (zn) for n ≥ 0. But the non-trivial ideals of k[[z]] arewell-ordered by ⊇, so we see (as in (7.5*)) that k[[z]] is Noetherian.

We show that each ideal of A is the contraction of an ideal of C[[z]], so A is Noetherian. Let f ∈ A ( C[[z]]be nonzero, and write f = z m g for g a unit of C[[z]]. The radius of convergence of g is the same as that of f , sog ∈A.2 The inverse g−1 of g in C[[z]] is also in A,3 so g−1 f = z m in A, showing ( f ) = (z m) again.

iii) The ring of power series in z with an infinite radius of convergence.This ring A is not Noetherian. Let fn =

∏∞j=n

1− z2

n2

for each n ≥ 1. It can be shown that fn ∈ A; in fact

πz f1 = sinπz. (If you like, replace the factors by 1− z2

22n , whose product more obviously converges.) Note that

the roots of 1− z2

n2 are z = ±n. Now each element of ( fn) vanishes at ±n, for limz→n | fn(z)| = 0, and if we haveg fn(n) 6= 0 for some g ∈ C[[z]], it follows limz→n |g (z)| =∞, and hence g has radius of convergence ≤ n and isnot in A. Now since fn+1| fn for each n, we have ( fn)⊆ ( fn+1), but fn+1 /∈ ( fn) since fn+1 does not vanish at±n. Thus( fn) is an infinite ascending series of ideals in A.

iv) The ring of polynomials in z whose first n derivatives vanish at the origin (n being a fixed integer).This ring A is Noetherian. We claim it is actually the subring A=C+ (zn+1)(C[z].4 Now C[zn+1]∼=C[z] is

Noetherian by (7.5), and the inclusionC[zn+1] ,→A makes A aC[zn+1]-module, finitely generated by 1, z, . . . , zn.Then (7.2) says A is Noetherian as well.

2 Write f (z) =∑

an zn+m and set c = limsupn→∞n+mp

|an | and R = 1/c . If |z | < R, then for some ε > 0, we have |z | < 1/(c + ε), whichimplies that for all sufficiently large n we have |an zn+m |< |an |(c+ε)−(n+m) ≤ c

c+εn+m . It follows that |an zn | ≤ 1

|z |mc

c+εn+m for n large enough,

so by comparison with the geometric series, g (z) converges.3 See the footnote to [10.8].4 Note first that the definition must be interpreted to involve the derivatives d

d z , . . . , d n

d zn ; the requirement specifically can’t include thatf = (d/d z)0( f )|z=0 = 0, because we want 1 in our ring. A really is a ring (actually a C-algebra), because derivatives are linear, and because by the

generalized Leibniz formula (see http://planetmath.org/encyclopedia/GeneralizedLeibnizRule.html) d n

d zn ( f g ) =∑n

j=0

nj

d n− j

d zn− j f ·d j

d z j g , so if the first n derivatives of f and g vanish at z = 0, then so do those of f g . Now d n

d zn (z m) = m · · · (m− n+ 1)z m−n , which is zero for

m ≤ n. If f =∑m

j=0 am z m , then d n

d zn

z=0( f ) =

∑mj=n a j j · · · ( j − n+ 1)0 j−n = n!an , which is zero just when an = 0. Thus the first n derivatives

of f vanish at 0 just if the coefficients a1, . . . , an are zero.

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Ex. 7.5 Chapter 7: Noetherian Rings

v) The ring of polynomials in z, w all of whose partial derivatives with respect to w vanish for z = 0.This ring A is not Noetherian. As in iv), the requirement must be on ∂ n

∂ wn

z=0for n strictly greater than 0 in

order that 1 ∈ A. The linearity of partial derivatives and the generalized Leibniz formula again show A really is aring. Let f ∈ A, and write f =

∑mj=0 p j (z)w

j for p j ∈ C[z]. If a j = p j (0) is the constant term, then ∂ n

∂ wn f

z=0=

∑mj=n a j j · · · ( j−n+1)w j−n , which vanishes identically in w just if all coefficients are zero. Then a j j · · · ( j−n+1) = 0

for all n > 0 and j ≥ n, so a j = 0 for all j > 0. Thus A=C[z]+ (z)C[w]. In particular, wn /∈ A, so we do not havezwn |zw m for m > n. Now let an = (zw, zw2, . . . , zwn) Ã A. Then an ( an+1 since zwn+1 /∈ an , so (an) is aninfinite ascending chain of ideals of A.

5. Let A be a Noetherian ring, B a finitely generated A-algebra, G a finite group of automorphisms of B, and BG the set ofall elements of B which are left fixed by every element of G. Show that BG is a finitely generated A-algebra.

Recall from [5.12] that B is integral over BG . Then (7.8) applied to the chain A⊆ BG ⊆ B says that BG is finitelygenerated as an A-algebra.

6. If a finitely generated ring is a field, then it is a finite field.[CAN THIS BE IMPROVED BY USING DGK, etc.?]Let k be our finitely generated field. We can without loss of generality assume 1 is part of the finite generating

set; then k is finitely generated over the subring A=Z · 1. Since k is a field, A must be Z or Fp . If A∼= Fp , then A isa field, and Zariski’s Lemma ((1.27.2*), (5.24), [5.18], (7.9)) shows k is a finite extension of Fp , so a finite field.

Otherwise A∼=Z, and we will derive a contradiction. k being a field, there is a subfield of k isomorphic toQ. Ask is finitely generated over this subfield, Zariski’s Lemma shows k is a finite (hence integral) extension of Q. Then(5.8) applied to the chain Z(Q⊆ k givesQ finitely generated over Z. ButQ is not finitely generated over Z.5

7. Let X be an affine algebraic variety given by a family of equations fα(t1, . . . , tn) = 0 (a ∈ I ) (Chapter 1, Exercise 27). Showthat there exists a finite subset I0 of I such that X is given by the equations fα(t1, . . . , tn) = 0 for a ∈ I0.

By (7.6) A= k[t1, . . . , tn] is Noetherian, so by (6.2) the ideal aÃ k[t1, . . . , tn] generated by fα : α ∈ I is finitelygenerated. Say its generators are g1, . . . , gn ∈ a. Then each gi is an A-linear combination of finitely many fαi , j

. NowI0 = αi , j : i = 1, . . . , n, j = 1, . . . , mi ⊆ I is a finite set, and fα : α ∈ I0 generates a. A point x ∈ kn satisfiesfα(x) = 0 for all α ∈ I0 just if f (x) = 0 for all f ∈ a just if fα(x) = 0 for all α ∈ I , so X is given by the vanishing of thefinite subset fα : α ∈ I0.

8. If A[x] is Noetherian, is A necessarily Noetherian?Yes. There is a canonical surjection A[x]A, so by (7.1), the quotient A is Noetherian if A[x] is.

9. Let A be a ring such that(1) for each maximal ideal m of A, the local ring Am is Noetherian;(2) for each x 6= 0 in A, the set of maximal ideals of A which contain x is finite.Show that A is Noetherian.

Let a Ã A be a non-zero ideal. By (2), for each x ∈ A the set Mx of maximal ideals m 3 x is finite, so the setM =

⋂

x∈a Mx of maximal ideals containing a is finite. For each m ∈ M , the extension S−1m a of a in Am is finitely

generated by (6.2) since (1) says Am is Noetherian. We may without loss of generality take these generators to beof the form x/1 for x ∈ a. Let am ⊆ a be the ideal of A finitely generated by these x. Now let z be an arbitrarynon-zero element of a, and let Nz be the finite set Mz \M . No n ∈ Nz contains a, so there is yn ∈ a\n. Nowlet b =

∑

m∈M am + (z) +∑

n∈Nz(yn) ⊆ a. This ideal is finitely generated. For m ∈ M , we have S−1

m a ⊆ S−1m b by

construction, so the two are equal. For maximal ideals m /∈ Mz , we have z ∈ Sm, so S−1m b = (1), and for n ∈ Nz we

have yn ∈ Sn, so S−1n b= (1). But if m /∈M , then S−1

m a= (1), and Max(A)\M = (Max(A)\Mz )∪Nz , so the localizationsof a and b are equal at all maximal ideals. This means the canonical injections (see (3.3)) Smb → Sma induced byb ,→ a are all surjective. (3.9) then says that the inclusion b ,→ a is surjective, so that b= a is finitely generated.

5 Suppose it were, say by a1/b1, . . . , an/bn for a j , b j ∈Z, or without loss of generality by 1/b j . Then if b =∏n

j=1 b j , thenQ=Z[1/b ]. Butthen if p ∈N is a prime not dividing b , we would have 1/p /∈Q, a contradiction.

96


10. Let M be a Noetherian A-module. Show that M [x] (Chapter 2, Exercise 6) is a Noetherian A[x]-module.This should be possible to prove in a way that specializes into the proof of Hilbert Basis Theorem (7.5) in the

case M =A. Let N be a submodule of M [x]; by (6.2), it will be enough to show N is finitely generated over A[x]. LetP ⊆M be the A-module consisting of its leading coefficients. As M is a Noetherian A-module, P is finitely generated,say by m1, . . . , mn . Let p1(x), . . . , pn(x) be elements of M [x] with leading coefficients mi , and let N ′ ⊆ N be theA[x]-submodule finitely generated by the pi (x). Say ri = deg(pi ) and r =maxn

i=1 ri .Suppose f (x) ∈ N has deg( f ) = p ≥ r and leading coefficient m ∈ P . Then there are ai in A such that m =

∑

ai mi , so∑

ai pi (x)xp−ri ∈N ′ is such that f ′(x) = f (x)−

∑

ai gi (x)xp−ri has deg( f ′)< m. By induction, there is

g ∈N ′ such that deg( f − g )< r .Write N ′′ =M +M x+ · · ·+M x r−1; then we have just shown N = (N ∩N ′′)+N ′. Now N ′′ is a finitely generated

A-module, so it is a Noetherian A-module by (6.5). Thus its submodule N ∩N ′′ is finitely generated over A by (6.2),and hence a fortiori finitely generated over A[x]. This shows N is finitely generated.

11. Let A be a ring such that each local ring Ap is Noetherian. Is A necessarily Noetherian?No. Let X be an infinite set and A=P (X ) the power set, viewed as a Boolean ring. We showed in [6.12] that A

is not Noetherian. Let p ∈ Spec(A), and a/s ∈Ap, for a ∈A and s ∈A\p. By the definition of a Boolean ring, a2 = aand s2 = s , so (a/s)2 = a2/s2 = a/s is idempotent. But Ap is a local ring, so by [1.12], a/s = 0 or 1. Thus Ap

∼= F2 isa field, hence surely Noetherian.

The other counterexample in [6.12] also works here. Recall that it is the subring A= k ·1+⊕

k of the countabledirect product

∏∞j=1 k, and it is not Noetherian. We claim that each localization at a prime is k, and hence Noetherian

(cf. [3.5]). Write kn for the nth direct summand of p0 :=⊕

kn ( A, and k0 := k · 1 ⊆ A. Recall that we defined en ∈kn ( B be the image of the 1 of kn and fn = 1−

∑

j≤n en , and the prime ideals of A are just p0 and pn := ( fn)+⊕

j 6=n k j

for n > 0. Their complements are Sn :=A\pn = pn+k×n . To find S−1n A, recall that for each n ≥ 0, A is generated as an

A-module by fn ∪ e j : j ≥ 1; thus there is an A-module surjection Mn =Afn ×⊕

j≥1 ke j →A. Since localizationis exact (3.3), S−1

n A is the image of the localization of the left-hand side under the induced map. For n ≥ 0, writeDn =Afn×

⊕

j 6=n ke j for the submodule of M mapping onto pn . It was shown in the course of proving [3.19.iv] thatlocalization distributes over arbitrary exact sums, so to compute S−1

n Mn , it will suffice to compute S−1n ke j for j 6= n,

S−1n ken , and S−1

n Afn . Since S0 contains each fn , and fn e j = 0 for j ≤ n, by [3.1] each S−10 e j = 0; on the other hand,

S−10 Af0 = (k

×0 )−1k0

∼= k0. Thus S−10 M ∼= k0

∼= k. For n > 0, since Sn contains each e j for j 6= n and e j el = 0 for j 6= l ,we have S−1

n ke j = 0, by [3.1], for j 6= n; since Sn contains en and en fn = 0, S−1n Afn = 0; but S−1

n kn = (k×n )−1kn

∼= kn ;so S−1

n M ∼= kn∼= k. Now for all n ≥ 0, S−1

n A is a quotient of S−1n A, so it is also isomorphic to k.

12. Let A be a ring and B a faithfully flat A-algebra (Chapter 3, Exercise 16). If B is Noetherian, show that A is Noetherian.Since B is faithfully flat over A we have aec = a for all aÃ A, so a 7→ ae is injective. Thus any infinite ascending

chain ⟨an⟩n∈N of ideals of A would give rise to an infinite ascending chain ⟨aen⟩n∈N of ideals of B .

13. Let f : A→ B be a ring homomorphism of finite type and let f ∗ : Spec(B)→ Spec(A) be the mapping associated with f.Show that the fibers of f ∗ are Noetherian subspaces of B [Spec(B), rather].

Recall (p. 30) that the homomorphism being of finite type means that B is finitely generated as an A-algebra (orequivalently, f (A)-algebra). Then B is a quotient of some polynomial ring C = A[t1, . . . , tm] (in particular, a C -algebra). Now recall from [3.21.iv] that the fiber ( f ∗)−1(p)≈ Spec

k(p)⊗AB

, where k(p) is the field Ap/pAp. By(2.14.iv) and (2.15), k(p)⊗AB ∼= k(p)⊗AC ⊗C B . But by [2.6] and induction,6 k(p)⊗AC = k(p)⊗AA[t1, . . . , tm]∼=k(p)[t1, . . . , tm], so the fiber is Spec

k(p)[t1, . . . , tm]⊗C B

. But C B is surjective, and (2.18) says tensor is rightexact, so k(p)[t1, . . . , tm]∼= k(p)⊗AC → k(p)⊗AB is surjective. (7.6) says k(p)[t1, . . . , tm] is Noetherian, so by (7.1),its quotient k(p)⊗AB is Noetherian, and by [6.8], this then has Noetherian spectrum.

Nullstellensatz, strong form

6 Assume M is an A-module, and inductively, M [x1, . . . , xn]∼=A[x1, . . . , xn]⊗AM . Pretty clearly A[x1, . . . , xn][y]∼=A[x1, . . . , xn , y]. Then

M [x1, . . . , xn , y][2.6]∼= A[y]⊗AM [x1, . . . , xn]

(2.14)∼= A[y]⊗AA[x1, . . . , xn]⊗AM[2.6]∼= A[x1, . . . , xn , y]⊗AM .

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14. Let k be an algebraically closed field, let A denote the polynomial ring k[t1, . . . , tn] and let a be an ideal in A. Let V bethe variety in kn defined by the ideal a, so that V is the set of all x = ⟨x1, . . . , xn⟩ ∈ kn such that f (x) = 0 for all f ∈ a.Let I (V ) be the ideal of V, i.e. the ideal of all polynomials g ∈A such that g (x) = 0 for all x ∈V. Then I (V ) = r (a).

Note that if we write V = Z(a), our goal is to show I Z(a) = r (a).By item 4 of [1.29], a⊆ I Z(a), and by item 9, I Z(a) = r

I Z(a)

, so r (a)⊆ I Z(a).On the other hand, suppose f /∈ r (a); we show f /∈ I Z(a). As f /∈ r (a), there is a prime p containing a but not

f , so g := f 6= 0 in the integral domain B = A/p. If C = Bg = B[1/g ], then by (1.3) C contains a maximal ideal n,and C/n is a field. Now C/n is a k-algebra, finitely generated over k by 1/g and the images of the ti . By Zariski’sLemma ((1.27.2*), (5.24), [5.18], (7.9)) C/n is a finite algebraic extension of k, and thus, since k is algebraically closed,isomorphic to k. Then the surjective k-algebra homomorphism

φ : AA/aA/p= B Bg =C C/n∼−→ k

has kernel m a maximal ideal of A containing a. By [1.27], m is of the form mx = (t1 − x1, . . . , tn − xn) for somex = ⟨x1, . . . , xn⟩ ∈ kn , so φ(t j ) = x j and φ : A→ k is the “evaluate at x” map h 7→ h(x). Since φ(a) = 0, we havex ∈ Z(a). Since g = f is a unit in C , its image remains a unit in k, so g (x) =φ(g ) 6= 0. It follows that g /∈ I Z(a).

Here is the classic proof of the Strong Nullstellensatz from the Weak ([5.17], cf. [5.18], [1.27], (5.24), (7.9)) bywhat is called the “Rabinowitsch trick.”7

Let k, A, and aÃA be as above, and suppose g ∈ I Z(a). If y now is a new indeterminate, consider the polynomialring A[y]. The polynomial 1− yg is 1 on the set Z(ae ) ⊆ kn+1, so that ae + (1− yg ) vanishes nowhere and henceby the Weak Nullstellensatz is the ideal (1) of A[y]. Then there are finitely many fi ∈ a and hi (y), h ′(y) ∈A[y] suchthat

1=∑

fi hi (y)+ (1− yg )h ′(y).

Under the A-algebra homomorphism A[y]→Ag ( k(t ) taking y 7→ 1/g , this equation is mapped to

1=∑

fi hi (1/g )+

1−gg

h ′(1/g ) =∑

fiHi

g m,

for some Hi ∈A and m =maxidegy hi. Multiplying through by g m shows g ∈ r (a).

To use the theory of Jacobson rings instead to prove the Strong Nullstellensatz, detouring past the Weak Null-stellensatz but not Zariski’s Lemma, proceed as follows.

Recall from [5.24] that A= k[t ] is a Jacobson ring and from [5.23] that a prime of A is an intersection of maximalideals. Since by [1.9] a radical ideal is an intersection of the primes containing it, it follows that a radical ideal in Ais the intersection of the maximal ideals containing it. Since I Z(a) is radical by item 9 of [1.29], it follows it is theintersection of the maximal ideals containing it, so it only remains to show each maximal ideal containing a alsocontains I Z(a). Taking X = x in item 0 of [1.27], we have x ∈ Z(a) ⇐⇒ a ⊆ mx , and taking X = Z(a) initem 8, we have x ∈ Z(a) ⇐⇒ I Z(a) ⊆ mx . But by the result of [1.27], these are the only maximal ideals of A, sor (a) = I Z(a).

For further proofs of the Nullstellensatz, see this discussion: http://mathoverflow.net/questions/15226/elementary-interesting-proofs-of-the-nullstellensatz. For numerous relatives, see Chapter 11 of PeteL. Clark’s notes http://math.uga.edu/~pete/integral.pdf.

15. Let A be a Noetherian local ring, m its maximal ideal and k its residue field, and let M be a finitely generated A-module.Then the following are equivalent:i) M is free;ii) M is flat;iii) the mapping of m⊗M into A⊗M is injective;iv) TorA

1 (k , M ) = 0.i) =⇒ ii): By (2.14.iv), A is a flat A-module. By [2.4], a direct sum of flat modules is flat, so a free A-module is

flat.7 This originated in the influential one-page paper [Rabinowitsch]. Just who Rabinowitsch was is an interesting question; it appears that he

later moved to the United States and became the influential mathematical physicist George Yuri Rainich. See also [MOPseud], [Mollin, p. 154],[Nark, p. 38].

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http://en.wikipedia.org/wiki/George_Yuri_Rainich


ii) =⇒ iii): This follows from (2.19) defining flatness.iii) =⇒ iv): We have a short exact sequence 0→m⊗M → A⊗M → k ⊗M → 0, implying Tor1(k , M ) = 0 by

the Tor exact sequence.iv) =⇒ i): Let x1, . . . , xn be such that their images form a basis of the finite-dimensional k-vector space k⊗AM ∼=

M/mM ; by (2.8), they generate M . Then there is an A-linear surjection An M taking e j 7→ x j , say with kernel N ,yielding a short exact sequence 0→ N → An → M → 0 of A-modules. Tensoring with k we have a Tor sequence(using “ii) ⇐⇒ iv)”) Tor1(k , M ) = 0→ k⊗N → kn→ k⊗M → 0. Since dimk kn = n = dimk (k⊗M ), linear algebra(or (6.9)) gives dimk (k⊗N ) = 0, so k⊗N = 0. As A is Noetherian, An is a Noetherian A-module by (6.5), and sinceN ⊆ An , by (6.2) N is finitely generated. k being finitely generated as well, [2.3] shows either k = 0 or N = 0. Byassumption, A 6=m, so k 6= 0. Therefore N = 0, and the map An→M is an isomorphism.8

16. Let A be a Noetherian ring, M a finitely generated A-module. Then the following are equivalent:i) M is a flat A-module;ii) Mp is a free Ap-module, for all prime ideals p;iii) Mm is a free Am-module, for all maximal ideals m.

In other words, flat = locally free.Since each Ap is local and each Mp finitely generated over Ap, [7.15] says Mp is free if and only if it is flat. The

equivalence then follows from (3.10).

17. Let A be a ring and M a Noetherian A-module. Show (by imitating the proofs of (7.11) and (7.12)) that every submoduleN of M has a primary decomposition.

Call a submodule M ⊆N irreducible if it is not an intersection of two proper supermodules; that is, for P1, P2 ⊆Msubmodules, we have N = P1 ∩ P2 =⇒ P1 =N or P2 =N .

If a submodule N ⊆M can be written as a finite intersection⋂n

j=1 P j of irreducible modules P j ⊆M , we call thisexpression an irreducible decomposition of N .

Lemma 7.11*. In a Noetherian A-module M , every submodule has an irreducible decomposition.SupposeΣ is the set of submodules of M not admitting an irreducible decomposition, and suppose for a contradic-

tion that Σ 6=∅. Then as M is Noetherian, Σ contains a maximal element N . Then N =⋂

N is not an irreducibledecomposition, by assumption, so N is not irreducible, and we can write it as N = P ∩ P ′, where P, P ′ ⊆ M aresubmodules strictly containing N . Then P and P ′ do admit irreducible decompositions P =

⋂

Q j and P ′ =⋂

Q ′j ,so N =

⋂

Q j ∩⋂

Q ′j is an irreducible decomposition, contradicting N ∈Σ. Thus Σ=∅.

Lemma 7.12*. In a Noetherian A-module M , every irreducible submodule is primary.If Q ⊆ M is a submodule, then N = M/Q is also Noetherian by Prop. 6.3. If Q were irreducible, then by the

correspondence of p. 18, the zero submodule of N = M/Q would be irreducible. We will show that if Q is notprimary, then 0⊆M/Q is reducible, so Q is reducible.

Let x ∈ A be a zero-divisor of N that is not nilpotent on N . The submodules 0 ⊆ (0 : x) ⊆ (0 : x2) ⊆ · · · ⊆ Nform an increasing chain; since N is Noetherian, the chain stabilizes at some P = (0 : x p ) = (0 : x p+1). Since x isnot nilpotent on N , we have P 6= N , so there is some n ∈ N \P . Then x p n 6= 0, so (x p )n 6= 0. If n′ ∈ (x p )n ∩ P ,we have an expression n′ = ax p n ∈ P , and multiplying by x gives 0 = ax p+1n. Then an ∈ (0 : x p+1) = (0 : x p ), son′ = ax p n = 0. Thus (x p )n ∩ P = 0. Since x is a zero-divisor of N , we have P 6= 0, showing 0⊆N is reducible.

Thus in an Noetherian A-module M , every submodule has an irreducible decomposition, and this is a primarydecomposition.

8 Here are two extra implications we don’t need.iv) =⇒ iii): We have a short exact sequence 0→ m→ A→ k → 0, giving a Tor exact sequence containing the fragment 0 = Tor1(k , M )→

m⊗M →A⊗M . This shows m⊗M →A⊗M is injective.iii) =⇒ ii): Let a be a finitely generated ideal of A. We have a short exact sequence 0→ a→ m→ m/a→ 0 of A-modules, whose Tor exact

sequence includes Tor1(k , M )→ a⊗M →m⊗M . Since iii) ⇐⇒ iv), the first term is zero; it follows that a⊗M m⊗M A⊗M is injective.Now the short exact sequence 0→ a→ A→ A/a→ 0 gives rise to a Tor exact sequence containing Tor1(A/a, M )→ a⊗M → A⊗M ; but we’vejust seen the kernel of the second map is zero, so Tor1(A/a, M ) = 0. By [2.26], then, M is flat.

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18. Let A be a Noetherian ring, p a prime ideal of A, and M a finitely generated A-module. Show that the following areequivalent:i) p belongs to 0 in M ;ii) there exists x ∈M such that Ann(x) = p;iii) there exists a submodule of M isomorphic to A/p.

ii) ⇐⇒ iii): Let x ∈M . Then the map AAx : a 7→ ax has kernel Ann(x), so induces an A-module isomorphismA/Ann(x) ∼−→ Ax ⊆ M . Then for any ideal a Ã A, there is a (cyclic) submodule isomorphic to A/a if and only ifa=Ann(x) for some x ∈M .

ii) =⇒ i): By [7.17], 0⊆M is decomposable. The primes belonging to 0 are those among the r (0 : y) = r

Ann(y)

for y ∈M by (4.5*) in [4.22]. Thus p=Ann(x) belongs to 0.i) =⇒ ii):9 By [7.17], 0 ⊆ M is decomposable, so let N1 ∩ · · · ∩Nn be an irredundant primary decomposition.

Let N be one of the N j , and N ′ the intersection of all the others. We will show will suffice to show p = rM (N ) =Ann(x) for some x ∈ M . Since the decomposition is irredundant, N ′ 6⊆ N , so we may find (and fix) a y ∈ N ′\N . Ifa1, . . . , am generate p j (possible by (6.2) since A is Noetherian) then there is for each a j some minimal p j ≥ 1 such

that ap jp y ∈N . If p =max j p j , then we see pp y ⊆N and pp−1y 6⊆N . Let x ∈ pp−1y\N then. Since y ∈N ′, we then

have px ⊆ pp y ⊆N ∩N ′ = 0, so p⊆Ann(x). On the other hand, if a ∈A is such that ax ∈N (hence = 0), then a isa zero-divisor of M/N ; as N is primary, this means a is nilpotent on M/N . But this says exactly that a ∈ rM (N ) = p.Thus p=Ann(x).10

Deduce that there exists a chain of submodules

0=M0 (M1 ( · · ·(Mr =M

such that each quotient Mi/Mi−1 is of the form A/pi , where pi is a prime ideal of A.Since M is Noetherian, 0 is decomposable by [7.17], so some p1 ∈ Spec(A) belongs to 0, and the above gives a

submodule M1∼= A/p1 of M . Assume inductively that we’ve found a chain 0= M0 ( · · ·( Mn ⊆ M . If Mn = M , we

are done; otherwise, M/Mn is Noetherian, so its submodule 0 is decomposable by [7.17]. Let pn+1 belong to it; thenthere is a submodule Nn+1 ⊆ M/Mn such that Nn+1

∼= A/pn+1. If Mn+1 ⊆ M is its pre-image under M M/Mn , wehave Mn+1/Mn

∼= Nn+1∼= A/pn+1. Since M is Noetherian, this process cannot create an infinite ascending chain, so

there is some Mr such that we cannot find an Mr+1. But we have shown that this only happens if Mr =M .

19. Let a be an ideal in a Noetherian ring A. Let

a=r⋂

i=1

bi =s⋂

j=1

c j

be two minimal decompositions of a as intersections of irreducible ideals. Prove that r = s and that (possibly afterre-indexing the c j ) r (bi ) = r (ci ) for all i .

State and prove an analogous result for modules.Since an ideal of A is just an A-submodule of A, a primary ideal of A is exactly a primary submodule of A (p.

50), an irreducible ideal of A is an irreducible submodule of A, rA(a) = r (a), and A is a Noetherian ring just if it isNoetherian as an A-module, it will be enough to prove the result in the more general case of a submodule N of aNoetherian A-module M .

We first prove the book’s hint: if⋂r

i=1 Pi =⋂s

j=1 Q j are minimal irreducible decompositions of N ⊆ M , andP ′k :=

⋂

i 6=k Pi for each k ∈ 1, . . . , r , then N equals one of the N j := P ′k ∩Q j . Note that all these modules containN , so it will be enough to show some N j ⊆ N .11 Surely since

⋂sj=1 Q j = N in M , we also have

⋂

j N j = N . Writeπi : M M/Pi for the natural map, andφ= (π1, . . . , πr ) : M →

⊕

i M/Pi for the induced map to the product. Nowker(φ) =

⋂

i Pi = N . For each j and each i 6= k, we have N j ⊆ P ′k ⊆ Pi , so πi (N j ) = 0. Thus the only potentially

9 Stolen from http://math.uiuc.edu/~r-ash/ComAlg/ComAlg1.pdf10 I worked for a while on another approach to this problem before turning to the experts, and this aborted effort went like this. By (4.5*) from

[4.22], p ∈ Spec(A) belongs to 0 just if (0 : x) : x ∈A & r (0 : x) = p is nonempty. A being Noetherian, this set contains some maximal elementq = (0 : x) = Ann(x). We suppose q 6= p and contradict maximality. If there is a ∈ p\q, then there is some minimal n ≥ 2 such that an x = 0,

but ax 6= 0. Consider (q : a) =

(0 : x) : a (1.12.iii)= (0 : ax). It contains (0 : x), properly since an−1 ∈ (0 : ax)\(0 : x). The proof will be concluded

if we can show r (0 : ax) = p. Unfortunately, I seem unable to do this. I wanted to say “Since a /∈ q, (4.4) says r (q : a) = r (q) = p, contradictingmaximality of q”; however, (4.4) requires as a hypothesis that q is primary, and it’s not clear to me this must be the case.

11 http://mathoverflow.net/questions/12322/atiyah-macdonald-exercise-7-19-decomposition-using-irreducible-ideals

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non-zero coordinate of an element of the module φ(N j ) is the j th, and it follows that these j th coordinates make upa submodule O j = πk (N j ) of M/Pk . Then 0 = φ

⋂

N j

=⋂

O j ×∏

i 6=k0, so⋂

O j = 0 ⊆ M/Pk . As Pk ⊆ Mis irreducible, 0 ⊆ M/Pk is irreducible, meaning for some j = j (k) we have O j = 0. Then πi (N j (k)) = 0 for all i ,including k, so N j (k) ⊆N .12

If we define 1Pi = Pi for i 6= k and 1Pk =Q j (k), we have another irreducible decomposition of N . Applying theprevious result to the 1Pi and the Q j , we can then replace another of the Pi by another Q j . Repeating this processfor k = 1, . . . , r , we eventually get an expression Q j (1)∩· · ·∩Q j (r ) = 0. Since we postulated irredundancy for the Q jdecomposition, it follows i 7→ j (i) is surjective and s ≤ r . A symmetric argument switching the roles of Pi and Q jwill then show r ≤ s , so we see r and s are equal.

Fact 7.A*. Any two minimal irreducible decompositions of a submodule have the same number of components.

It remains to show we may reorder the Qi so that rM (Qi ) = rM (Pi ) for all i ∈ 1, . . . , s. Recall from Lemma7.12* of [7.17] that, since M is Noetherian, each irreducible module is primary, and from (4.3*) of [4.21], that an in-tersection of p-primary modules is p-primary; thus by collecting terms with the same radical, we obtain irredundantprimary decompositions M1∩ · · ·∩Mm =N1∩ · · ·∩Nn of N ⊆M . By (4.5*) of [4.22], the sets rM (M1), . . . , r (Mm)and rM (N1), . . . , r (Nm) ⊆ Spec(A) are equal, and so we may renumber them so that pi = r (Mi ) = r (Ni ). We mayfurther reorder them, since there are only finitely many, so that each set S j := p1, p2, . . . , p j ⊆ Spec(A) is isolated.Suppose that Mi is the intersection of mi different P j , and Ni is the intersection of ni different Q j . It will thensuffice to show that mi = ni for each i . By Thm. 4.10* of [4.23], applied to the isolated set S1, we see M1 = N1.Irredundancy of the irreducible decompositions of N show the relevant P j and Q j each give a minimal irreducibledecomposition of M1, and then Fact 7.A above shows that m1 = n1. Assume inductively that we have shown m j = n j

for all j < i . Thm. 4.10* of [4.23], applied to the isolated set Si , shows that N ′i :=⋂i

j=1 M j =⋂i

j=1 N j . Irredundancyof the irreducible decompositions of N again show the relevant P j and Q j give a minimal irreducible decompositionof N ′i , and Fact 7.A shows that

∑ij=1 m j =

∑ij=1 n j . By inductive assumption m j = n j for all j < i , so subtracting

these off, mi = ni , and we are done.

20. Let X be a topological space and letF be the smallest collection of subsets of X which contains all open subsets of X andis closed with respect to the formation of finite intersections and complements.i) Show that a subset E of X belongs toF if and only if E is a finite union of sets of the form U ∩C , where U is open andC is closed.

Let G be the collection of finite unions of sets U ∩C , for U open and C closed.First we show G ⊆ F . Each open set is in F , and taking complements, each closed set is in F . Taking inter-

sections, each U ∩C ∈ G for U open and C closed. ButF is closed under finite unions, for by De Morgan’s laws,⋃

Si =X \⋂

(X \Si ), andF is assumed to be closed under complement and finite intersection.Now we showF ⊆G by showing G satisfies the properties (except “smallest”) postulated ofF .

• Taking C =X, each open U ⊆X is in G .

• Finite intersections: It suffices to prove this for binary intersections. Let S =⋃

(Ui ∩Ci ) and S ′ =⋃

(U ′j ∩C ′j )be inG . Then S∩S ′ =

⋃

i (Ui∩Ci )∩⋃

j (U′j ∩C ′j ) =

⋃

i , j Ui∩U ′j ∩Ci∩C ′j by distributivity; since each Ui∩U ′jis open and each Ci ∩C ′j is closed, S ∩ S ′ ∈G .

• Complements: If S =⋃n

i=1(Ui ∩ Ci ) ∈ G , then De Morgan’s laws give X \S = X \⋃

(Ui ∩ Ci ) =⋂

X \(Ui ∩Ci ) =

⋂

[(X \Ui )∪ (X \Ci )]. Now Si1 = X \Ci is open and Si2 = X \Ui is closed. Let F be the set ofall functions 1, . . . , n → 1, 2, and for f ∈ F write S f =

⋂ni=1 Si , f (i). Then each S f is a finite intersection

of open and closed sets, hence an intersection of one closed set and one open set. Distributivity then givesX \S =

⋂

[(X \Ui )∪ (X \Ci )] =⋃

f∈F S f ∈G .

12 I initially attempted a simpler argument as follows. Consider the projection π : M → M/Pk . Since⋂

Q j =N , we have π⋂

Q j

= 0. SincePk ⊆ M is irreducible, 0 ⊆ M/Pk is irreducible, and one of the π(Q j ) = Q j /Pk = 0. This doesn’t seem to work as stated, because there’s noguarantee that Q j ⊇ Pk , so that π(Q j ) is a submodule.

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ii) Suppose that X is irreducible and let E ∈ F . Show that E is dense in X (i.e., that E = X ) if and only if E contains anon-empty open set in X.

If E contains a non-empty open set U ⊆ X, then X = U ⊆ E by [1.19]. Now suppose E =⋃

(Ui ∩Ci ) ∈ F isdense in X. Recalling that closure distributes over finite unions,13 we see X = E =

⋃

Ui ∩Ci . Recalling from [6.7]that an irreducible space is not a union of finitely many proper closed subspaces, for some i we have Ui ∩Ci = X.Now Ui ∩Ci ⊆Ui ∩Ci so Ui =X =Ci =Ci . Then E contains the open set Ui =Ui ∩Ci , which is non-empty sinceit is dense.

21. Let X be a Noetherian topological space (Chapter 6, Exercise 5) and let E ⊆ X. Show that E ∈ F if and only if, for eachirreducible closed set X0 ⊆X, either E ∩X0 6=X0 or else E∩X0 contains a non-empty open subset of X0. The sets belongingtoF are called the constructible subsets of X.

If E ∈ F , then the previous [7.20] says that either E ∩X0 contains a non-empty open set of X0 or is not dense.But it is not dense precisely if its closure in X0 is not all of X0.

We prove the other direction by contraposition. Suppose E /∈F . Then trivially X ∩ E /∈F , so the set of closedC ⊆X with C∩E /∈F is non-empty. Since X is Noetherian, there are minimal such sets; let X0 be one. If C , C ′ (X0are closed, then by minimality, C ∩E and C ′∩E are inF , soF also contains their union (C ∪C ′)∩E . Apparently,then, we cannot have C ∪C ′ =X0, showing X0 is irreducible. We will show that E ∩X0 is dense in X0, yet containsno nonempty open subset of X0.

Write F for the closure of E ∩X0 in X0. If F is a proper subset of X0, then by minimality, E ∩ F ∈ F . NowF = E ∩X0,14 so E ∩X0 = E ∩ E ∩X0 = E ∩ F ∈F , contrary to assumption. Therefore F =X0.

If E ∩X0 contained a nonempty open subset U of X0, then either X0 = U = E ∩X0, contradicting E ∩X0 /∈F ,or ∅ 6= C := X0\U ( X0. By the definition of the subspace topology, C is closed in X and U = V ∩X0 for someopen V ⊆ X. Then E ∩U = U = V ∩X0 ∈ F , and by minimality of X0 we have C ∩ E ∈ F . Since U ∪C = X0,we then have E ∩X0 = (E ∩C )∪ (E ∩U ) a union of elements of F , hence in F itself, contrary to assumption. Itfollows that E ∩X0 contains no nonempty open subset of X0.

22. Let X be a Noetherian topological space and let E be a subset of X. Show that E is open in X if and only if, for eachirreducible closed subset X0 in X, either E ∩X0 =∅ or else E ∩X0 contains a non-empty open subset of X0.

Suppose first E ⊆X is open. Then for any subspace X0 ⊆X, by definition E ∩X0 is an open subset of X0; eitherit is empty, or it is not.

We prove the other direction by contraposition. Now suppose E ⊆ X is not open; then E ∩X is trivially notopen, so the collection of closed subsets C ⊆ X with C ∩ E not open in C is non-empty. As X is Noetherian, thereare minimal elements of this collection; let X0 be one. If C , C ′ (X0, then by minimality C ∩E and C ′∩E are openin X0, so their union (C ∪C ′) ∩ E is open in X0. It follows that C ∪C ′ 6= X0, so X0 is irreducible. We will showE ∩X0 is non-empty, yet contains no non-empty open subset of X0.

If E ∩X0 were ∅, this intersection would be open, contrary to assumption, so the two sets do meet. Suppose fora contradiction that we can find a non-empty open subset U ⊆X0∩E of X0. Write C =X0\U ; since U is nonempty,C (X0, so by minimality of X0 we have C ∩E open in C . By the definition of the subspace topology, then, there isan open V ⊆X with C ∩ E =V ∩X0. Then X0 ∩ E = (U ∩ E)∪ (C ∩ E) =U ∪ (V ∩X0) is a union of open subsetsof X0, hence open in X0, contrary to assumption.

23. Let A be a Noetherian ring, f : A→ B a ring homomorphism of finite type (so that B is Noetherian). Let X = Spec(A),Y = Spec(B) and let f ∗ : Y →X be the mapping associated with f . Then the image under f ∗ of a constructible subset Eof Y is a constructible subset of X .

To see B is Noetherian, recall from p. 30 that being of finite type means that B is finitely generated as an A-algebra,hence a quotient of a polynomial ring over A. Then (7.5) and (7.1) show B is Noetherian.

Since f ∗⋃

Si

=⋃

f ∗(Si ), it will suffice to consider E of the form U ∩C for U ⊆ Y open and C ⊆ Y closed.An open subset U ⊆ Y is a union of basic open sets Yg for g ∈ B ([1.17]). By [6.6], U is compact, so it is a union offinitely many Ygi

. Then U ∩C =⋃

(Ygi∩C ), so it will suffice to verify the statement for E = Yg ∩C . Since a closed

subset C ⊆ Y is V (b) for some ideal bÃ B ([1.15]), we may assume E = Yg ∩V (b).

13 A⊆A∪B , so A⊆A∪B , and similarly B ⊆A∪B , so A∪B ⊆A∪B . On the other hand, A∪B ⊆A∪B , and the latter is closed, so A∪B ⊆A∪B .14 In general, if A⊆ Y ⊆X, the closure B of A in Y is equal to A∩Y . On the one hand, the latter is closed in Y and contains A, so B ⊆A∩Y .

On the other hand, if x ∈A∩Y , then x ∈ Y and every neighborhood U 3 x contains some point of A⊆ Y , so every neighborhood U ∩Y ⊆ Yof x meets A, and thus x ∈ B .

102


Now by [1.21.iv], V (b) ≈ Spec(B/b). Writing π : B B/b and Spec(B/b) = Z , [1.21.i] says for q ∈ V (b) wehave q= π∗(q/b) ∈ Yg ⇐⇒ π(q) = q/b ∈ (π∗)−1(Yg ) = Zπ(g ), so E =V (b)∩Yg = π

∗(Zπ(g )). Replacing f : A→ Bby A→ B B/b, B by B/b, and Yg by Zπ(g ), and noting B/b is also of finite type over A, we may assume that E ⊆ Bis a basic open set.

If φg : B → Bg is the canonical map, then E = Yg =φ∗g

Spec(Bg )

. Since we have a B -algebra surjection B[t ]Bg taking t 7→ 1/g , it follows Bg is of finite type over A, and we may replace B by Bg , f byφg f , and E by Spec(Bg ).Now we have only to show that given a map f : A→ B of finite type, with A Noetherian, f ∗(Y ) is constructible.15

We will attempt to use [7.21]. Let an irreducible closed set X0 ⊆ X be given. By the proof of [1.20.iv], X0 is ofthe form V (p) for some prime p ∈ X .16 Write F = f ∗(Y )∩X0 for the set we want to prove constructible. We havep′ ∈ f ∗(Y )∩X0 if and only if there is q′ ∈ Y∩( f ∗)−1(X0) = ( f

∗)−1(X0) such that p′ = f ∗(q′); then p′ ∈ f ∗

( f ∗)−1(X0)

.Thus F = f ∗

( f ∗)−1(X0)

. By [1.21.ii], ( f ∗)−1

V (p)

=V (pB) is the pre-image of F and by [3.21.iii], f ∗ “restricts”to a map (which we call f ∗ again) Spec(B/pB) → Spec(A/p), whose image we want to show is constructible; therings are still Noetherian, with the map still of finite type. So we may assume A is a Noetherian integral domain andf : A→ B is of finite type, and we want to show f ∗(Y ) is constructible. Since B is then Noetherian, Y is a Noetherianspace, so by [6.7], it is a union of finitely many irreducible components Y1, . . . , Yn . Now by [7.21], it is enough toshow either f ∗(Y ) is not dense or contains a non-empty open set. If f ∗(Y ) is dense, then X = f ∗(Y ) =

⋃

f ∗(Yj ),so by irreducibility of X , we have some f ∗(Yj ) dense17; we want to show this f ∗(Yj ) contains a non-empty openset. Write Yj = V (q) for a prime ideal q Ã B ([1.20.iv]); then f ∗(Yj ) is the image of the map on spectra inducedby the composition A→ B B/q. Replacing B with B/q and f with this composition, we may assume A and Bare Noetherian integral domains. Since f ∗(Y ) is dense in X , by [1.21.v], we must have ker( f ) ⊆N(A) = 0, so f isinjective. Now we are in the situation of [5.21]. There exists a nonzero s ∈A such that given an algebraically closedfield Ω and homomorphism φ : A→ Ω such that φ(s) 6= 0, we can extend φ to a homomorphism B → Ω. Since theimages of these maps are subrings of a field, hence integral domains, it follows that their kernels p, q are prime, withp= A∩ q= f ∗(q). We have φ(s) 6= 0 ⇐⇒ s /∈ ker(φ) = p, so for all p ∈ Xs we have p ∈ f ∗(Y ), so that Xs ⊆ f ∗(Y )is an open subset. But since A is an integral domain, s is not nilpotent, and thus by [1.17.ii], Xs 6=∅.

The previous solution followed the book’s breadcrumb trail; the following one is adapted from another solution15 This following paragraph is an approach I once thought would work, but does not. I include it because I spent a good deal of time on it and

its failure, at least to me, seemed somewhat subtle.Since f is of finite type, it can be factored as π ι, where ι : A ,→ A[x1, . . . , xn] is the canonical inclusion into a polynomial ring and

π : A[x1, . . . , xn] B is a quotient map. Now f ∗(Y ) = ι∗

π∗(Y )

, where π∗(Y ) =V

ker(π)

by [1.21.iv]. If we vary B and f , this set varies overall closed subsets of Spec(A[x1, . . . , xn]); thus we may assume f : A ,→ B =A[x1, . . . , xn] and E =V (b) for some bÃ B .

By [1.21.iii], we have f ∗(E) = V (bc ), so certainly f ∗(E) ⊆ V (bc ). It would be nice if we could prove the reverse inclusion. (If that wereso, however, it would follow that Spec(A[x1, . . . , xn])→ Spec(A) is always a closed map, which we show below is not the case. By [6.11], thisis equivalent to f having the going-up property, which we also show is not so in general.) To attempt this, let p ∈ V (bc ) be given; we wishto find a prime q ∈ V (b) such that A∩ q = p. By [3.21.iii], f ∗ “restricts” to a map Spec(B/pe ) → Spec(A/p) that we will also call f ∗. Sincepe = p[x1, . . . , xn], the domain of the new f ∗ is the integral domain (A/p)[x1, . . . , xn]; what we now need is a prime q of (A/p)[x1, . . . , xn]containing the extension

c := be Ã (A/p)[x1, . . . , xn]. (7.1)Replacing A with A/p and b with c, we may assume A is a Noetherian integral domain, b is an ideal of B =A[x1, . . . , xn] such that A∩b= 0, andwe need a prime q ∈V (b) such that A∩ q = 0. Write S = A\0 and K = S−1A for the field of fractions of A. Then S−1B = K[x1, . . . , xn], and[3.21.ii] gives a commutative diagram

(0)= Spec(K)

Spec

K[x1, . . . , xn]oooo

Spec(A) Spec

A[x1, . . . , xn]

.oooo

In K[x1, . . . , xn] there is certainly a maximal ideal n containing the extension S−1b = be , since b∩ S = ∅ implies, by (3.11.ii), that S−1b 6= (1).Then using (1.17.i), b⊆ bec ⊆ nc =: q, and following n both ways around the diagram, we see q∩A= n∩A= (0).

Now we have a “proof” that never requires the Noetherian hypothesis on A (or the finite type, either, seemingly). The argument breaks downat Eq. 7.1; it can easily happen that be = p, so there is no prime containing it. For an example, consider A = Z(p) and the ideal q = (p x − 1)in Z(p)[x] (http://bit.ly/S9wGoP). Then Z(p)[x]/q∼= Z(p)[1/p]∼=Q, so q is maximal, and hence q( Spec

Z(p)[x]

is closed by [1.18.i].But q ∩Z(p) = (0), and (0) is not closed in Spec

Z(p)

, since its closure is V

(0)

= (0), (p) ([1.8.ii]). Also, the extension (c above) of q inZ(p)/(p)Z(p) ∼=Z/pZ is (−1) = (1), so no prime contains it.

16 The next thing I wanted to do is as follows. I leave it to posterity to rescue it, if possible. Eq. 1.1, (1.18), and (1.8) give f ∗(Y ) =V⋂

f ∗(Y )

=

V

f⋂

Y )

=V

N(B)c

. Then the closure of f ∗(Y )∩V (p) in V (p) is f ∗(Y )∩V (p) =V

N(B)c

∩V (p) by the footnote to [7.21]. If it equals

V (p); then each prime p′ Ã A containing p also contains N(B)c ; in particular, p contains N(B)c . This should hypothetically help us find a basicopen subset Xh contained in f ∗(Y )∩V (p), but I do not know how.

17 This line from Yimu Yin’s solution: http://pitt.edu/~yimuyin/research/AandM/exercises07.pdf

103

http://bit.ly/S9wGoP



set online.18 We want to prove f ∗(Y ) is constructible, assuming only that f is of finite type and A Noetherian. Giventhat Y is Noetherian, and by [6.7] a union of finitely many irreducible components Y1, it will be enough to show eachf ∗(Y1) is constructible. Let q1 be a minimal prime of B such that V (q1) = Y1 ([1.20.iv]); then the image F = f ∗(Y1)is contained in V (qc ) by [1.21.iii], so f ∗ restricts to a map ([3.21.iii]) B1 = Spec(B/q1) → Spec(A/qc

1) = A1. Iff ∗(Y1) is constructible in X1 = Spec(A/qc

1), it will also be in X : if f ∗(Y ) is a union of sets U ′ ∩ C ′ for U ′ openand C ′ closed in X1, then there are U open and C closed in X such that U ′ = X1 ∩U and C ′ = X1 ∩C , and thenU ′∩C ′ =U∩(C∩X1). Note that A1 and B1 are integral domains and f1 : A1→ B1 is injective. As in the previous proof,by [5.21] there is s1 ∈A1 such that s1 /∈ p1 ∈X1 implies p1 ∈ f ∗1 (Y1); thus Xs1

⊆ f ∗1 (Y1) = F . Now s1 is in each primein V (s1) = X1\Xs1

, so we may consider ([3.21.iii] again) the restriction of f ∗1 to the map ( f ∗1 )−1

V (s1B)

→ V (s1)as induced by the homomorphism A′1 = A1/(s1) → B1/(s1)B1 = B ′1. By [6.5], Y1 is Noetherian, so by [6.7], it hasfinitely many irreducible components. Let Y2 be one, and q2 such that ([1.20.iv]) V (q2) = Y2. The image ( f ′1 )

∗(Y2)induced by f ′1 : A′1→ B ′1 is the same as (making identifications) the image f ∗2 (Y2) induced by f2 : A′1/q

c2→ B ′1/q. Again

this is an injection of integral domains, so we can find an open subset Xs2⊆ X2 contained in f ∗2 (Y2). Iterating, we

get a sequence of sets Xs j⊆ F , where each Xs j+1

is open in the closed set X j \Xs j. That means that there is an open

Uj+1 ⊆X j such that Uj+1 ∩ (X j \Xs j) =Xs j+1

. Thus Xs j+1∪Xs j

=Uj ∪Xs jis open in X j . Therefore Wn =

⋃nj=1 Xs j

isan increasing chain of open sets in F ; since X is Noetherian, so is F by [2.5], so this chain terminates in some Wn .

At each point where we chose an irreducible component in the above process, we could have instead chosen adifferent irreducible component, and obtained a different open subset W ⊆ F . If we do this for each possible chainof irreducible components, and take the union, we have obtained F as an open set, which is certainly constructible.

24. With the notation and hypotheses of Exercise 23, f ∗ is an open mapping ⇐⇒ f has the going-down-property (Chapter5, Exercise 10).

By [5.10.ii], if f : A→ B is any ring homomorphism such that f ∗ is open, then f has the going-down property.Conversely, suppose f has the going-down property, and let Ys be a basic open set ([1.17]) in Y = Spec(B);

it is enough to show f ∗(Ys ) is open. By (3.11.iv), the canonical map B → Bs has the going-down property, andA→ B→ Bs is still of finite type, so replacing B with Bs and f with the composition, it is enough to show f ∗(Y ) isopen. Let X0 be an arbitrary irreducible closed subset of X . By [7.22], to show f ∗(Y ) is open it will suffice to showthat either it does not meet X0 or F = f ∗(Y )∩X0 contains a non-empty open subset of X0. Assume that q ∈ F . Thenif p′ ⊆ q is another prime, we also have p′ ∈ f ∗(Y ), by going-down. But [1.20.iv] tells us there is a prime p such thatX0 =V (p), and thus p ∈ F . Since p=V (p) =X0 ([1.18.ii]), we then have F dense in X0. By the previous problem,f ∗(Y ) is constructible, and so F is as well. [7.20.ii] then says that F contains a non-empty open set in X0.

25. Let A be Noetherian, f : A→ B of finite type and flat (i.e., B is flat as an A-module). Then f ∗ : Spec(B)→ Spec(A) is anopen mapping.

By [5.11], since f is flat, it has the going-down property. By [7.24], then, f ∗ is an open mapping.

Grothendieck groups26. Let A be a Noetherian ring and let F (A) denote the set of all isomorphism classes of finitely generated A-modules. Let

C be the free abelian group generated by F (A). With each short exact sequence 0 → M ′ → M → M ′′ → 0 of finitelygenerated A-modules we associate the element [M ′]− [M ]+[M ′′] of C , where [M ] is the isomorphism class of M , etc. LetD be the subgroup of C generated by these elements, for all short exact sequences. The quotient group C/D is called theGrothendieck group of A, and is denoted by K(A). If M is a finitely generated A-module, let γ (M ), or γA(M ), denote theimage of [M ] in K(A).

Before proceeding, we establish some additional notation. Let F (A) be the class of all finitely generated A-modules, [−]A :F (A) F (A) the function taking a module to its isomorphism class, iA : F (A) ,→C (A) the canonicalinclusion of generators, and πA : C (A)K(A) =C (A)/D(A) the quotient map. Note that γA=πA iA [−]A.

Since only one ring occurs in the discussion up to part iv), we will until then mostly suppress mention of A.19

18 http://pitt.edu/~yimuyin/research/AandM/exercises07.pdf19 As an aside, note, although it’s not strictly necessary for us to do so, that if 0→ N → M → P → 0 is a short exact sequence of A-modules

and N and P are finitely generated, then so is M , by [2.9]. It’s also true that if M is finitely generated, then it is Noetherian by (6.5), so N and Pare finitely generated. Again, generators of D are defined to be linear combinations of finitely generated classes, but these closure properties aresomehow reassuring.

104



i) Show that K(A) has the following universal property: for each additive function λ on the (proper) class of finitelygenerated A-modules, with values in an abelian group G, there exists a unique homomorphism λ0 : K(A)→ G suchthat λ(M ) = λ0

γ (M )

for all M .Since for any short exact sequence 0→N → M → P → 0 of finitely generated A-modules we have [N ]− [M ]+

[P ] ∈D = ker(π), its image under π is 0=π

[N ]

−π

[M ]

+π

[P ]

= γ (N )−γ (M )+γ (P ), so γ :F →K is itselfan additive function (p. 23). We will show a little more than the claim: not only is it the case that for each additivefunction λ :F →G is there a unique homomorphism λ0 : K→G such that λ= λ0 γ , but to each homomorphismλ0 : K →G corresponds a unique additive function λ= λ0 γ :F →G, so the correspondence λ↔ λ0 is bijective.In other words, γ is a universal additive function onF .

For the correspondence λ0 7→ λ, recall that γ is additive, so for any short exact sequence 0→N →M → P → 0 offinitely generated A-modules we have γ (N )− γ (M )+ γ (P ) = 0. It follows that for any homomorphism λ0 : K→G,if we define λ= λ0 γ , then λ(N )−λ(M )+λ(P ) = λ0

γ (N )− γ (M )+ γ (P )

= λ0(0) = 0, so that λ is additive.

For the correspondence λ 7→ λ0, given an additive map F → G, we find a homomorphismλ0 : K → G such that λ0 γ = λ, then show it is unique. First, to see λ descends (uniquely) to awell defined function on F , consider the very boring short exact sequence 0→ 0→ 0→ 0→ 0.By additivity, λ(0) = λ(0)− λ(0) + λ(0) = 0. If M ∼= N ∈ F , then there exists a short exactsequence 0→ M → N → 0→ 0, so by additivity, λ(M )− λ(N ) = λ(M )− λ(N ) + λ(0) = 0, andλ(M ) = λ(N ). Thus λ(M ) depends only on the isomorphism class of M and λ descends to a welldefined function λF : F →G with λF [−] = λ. Second, as C is the free abelian group on F , wehave a unique homomorphism λC : C →G such that λC i = λF . Third, given any short exactsequence 0→N →M → P → 0, we haveλC

[N ]−[M ]+[P ]

= λC

[N ]

−λC

[M ]

+λC

[P ]

=λ(N )−λ(M )+λ(P ) = 0, so D ⊆ ker(λC ) and λC descends to a homomorphism λ0 : K =C/D→G, with λ0 π= λC . As requested, λ0 γ = λ0 π i [−] = λC i [−] = λF [−] = λ.

F (A)

γ

[−]

λ

F (A) _i

λF

C (A)

π

λC

!!K(A)

λ0

// G

To see uniqueness, note that the [M ] ∈ C for M ∈ F are generators for C , so their images γ (M ) = π

[M ]

generate K = C/D . Thus if a partially defined function l0 : K → G satisfies l0

γ (M )

= λ(M ), there is at most oneway to extend l0 to a totally defined homomorphism λ0 : K→G.

ii) Show that K(A) is generated by the elements γ (A/p), where p is a prime ideal of A.Let M ∈ F , and recall from [7.18] (this is the first time we use that A is Noetherian) that there exists a chain

0= M0 ( M1 ( · · ·( Mr = M of submodules with successive quotients of the form A/p j for p j ∈ Spec(A). Thus wehave for 1≤ j ≤ r short exact sequences 0→ M j−1→ M j → A/p j → 0, showing that [M j−1]− [M j ]+ [A/p j ] ∈ D ,and so γ (M j ) = γ (M j−1)+ γ (A/p j ). Since γ (M0) = γ (0) = 0, it follows by induction that γ (Mn) =

∑

j≤n γ (A/p j ); inparticular, γ (M ) =

∑rj=1 γ (A/p j ).

iii) If A is a field, or more generally if A is a principal ideal domain, then K(A)∼=Z.First note that a field is a principal ideal domain (henceforth “PID”), with only the two ideals (0) and (1). Second,

note that a principal ideal domain A is Noetherian by (6.2), since every ideal is by definition principal, so a fortiorifinitely generated. By ii) above K is then generated by the elements γ (A/p)with p ∈ Spec(A). Let (a) 6= (0) be any idealof A; since A is an integral domain, x 7→ ax is injective, so we have a short exact sequence 0→ A→ A→ A/(a)→ 0of A-modules, meaning γ

A/(a)

= γ (A)−γ (A)+γ

A/(a)

= 0 in K . Since A is an integral domain, (0) is prime, andso γ

A/(0)

= γ (A) generates K . Thus K is a quotient of Z, and it remains to show it isn’t a proper quotient. To dothis, it suffices to define a surjective homomorphism KZ, and by i), it is enough to produce an additive functionF →Z with image N.20

Recall21 that any finitely generated module M over a PID A can be written as a finite direct sum Ar ⊕T (M ) forsome uniquely determined rank r = rkA M ∈ N, where T (M ) is the torsion submodule of M ([2.12]). λ := rkA hasimage N since λ(An) = n for all n ∈N. Write L for the field of fractions of A. For any nonzero m ∈ T (M ), say witha ∈Ann(m)\0, we have 1⊗m = (a/a)⊗m = (1/a)⊗am = (1/a)⊗0= 0 in L⊗M , so L⊗M ∼= L⊗Aλ(M ) ∼= Lλ(M )

by (2.14.iii,iv). Since by (3.6) L is a flat A-module, tensoring a short exact sequence 0→N →M → P → 0 of finitelygenerated A-modules with L gives rise to a short exact sequence 0→ Lλ(N )→ Lλ(M )→ Lλ(P )→ 0 of L-vector spaces.We know from linear algebra that λ(N )−λ(M )+λ(P ) = 0, so λ is additive.

20 I am not pleased with this proof; it seems intuitively obvious γ (A) is not torsion, so I think there should be a quicker argument.21 http://planetmath.org/encyclopedia/FinitelyGeneratedModulesOverAPrincipalIdealDomain.html

105

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iv) Let f : A→ B be a finite ring homomorphism. Show that restriction of scalars gives rise to a homomorphism f! : K(B)→K(A) such that f!

γB (N )

= γA(N ) for a B-module N . If g : B → C is another finite ring homomorphism, show that(g f )! = f! g!.If N is a finitely generated B -module, write N | f for the A-module obtained therefrom by restriction of scalars.

(2.16) says that then N | f is a finitely generated A-module, so we have a function | f :F (B)→F (A). An short exactsequence 0→N →M → P → 0 of B -modules remains exact viewed as a sequence of A-modules, so γA| f is an additivemap F (B)→ K(A). By the universal property of i), there is then a unique homomorphism f! : K(B)→ K(A) suchthat f! γB = γA | f .

Let M be a C -module, a ∈ A, and x ∈ M . Note that M |g ,

M |g

f , and M |g f have the same underlying abelian

group, so the last two will be equal A-modules if they have the same A-action. The element a · x of

M |g

f , by

definition, is the element f (a) · x in M |g , which in turn is g

f (a)

· x = (g f )(a) · x in M , which is a · x in M |g f .Thus

M |g

f =M |g f . Abstracting, we can write |g f = |g | f :F (C )→F (A). Now consider the diagram

F (A)

γA

F (B)| foo

γB

F (C )|goo

γC

K(A) K(B)

f!oo K(C ).g!

oo

Commutativity of the two squares implies commutativity of the outer rectangle: f! g!γC = f!γB |g = γA| f |g =γA |g f . Then f! g! is a homomorphism φ : K(C )→ K(A) satisfying φ γC = γA |g f . As (g f )! was defined tobe the unique homomorphism with this property, it follows that (g f )! = f! g!.

27. Let A be a Noetherian ring and let F1(A) be the set of all isomorphism classes of finitely generated flat A-modules. Repeatingthe construction of Exercise 26 we obtain a group K1(A). Let γ1(M ) denote the image of [M ] in K1(A).

Two asides follow.2223

i) Show that tensor product of modules over A induces a commutative ring structure on K1(A), such that γ1(M )·γ1(N ) =γ1(M ⊗N ). The identity element of this ring is γ1(A).First recall the notation we introduced in the proof of [7.26]. Let F1 ⊆ F be the class of all finitely generated

flat A-modules. Note F1 ⊆ F , so the map [−] : F F restricts to a map F1 F1. Let C1 be the free group onF1. Since F1 ⊆ F , there is a natural embedding C1 C , which we view as an inclusion. Write D1 ⊆ C1 for thesubgroup generated by [N ]− [M ] + [P ] for short exact sequences 0 → N → M → P → 0 of objects in F1, andπ1 : C1K1 =C1/D1 for the quotient map.

A tensor product of finitely generated modules is finitely generated for if M , N are generated by some finitelymany xi ∈M and y j ∈N , then M ⊗AN is generated by the finitely many xi ⊗ y j . Thus−⊗− is a mapF ×F →F .By [2.8.i], a tensor product of flat modules is flat, so −⊗− restricts to a mapF1×F1→F1. Given isomorphismsφ : M

∼−→M ′ and ψ : N∼−→N ′, we have an isomorphism φ⊗ψ : M ⊗N

∼−→M ′⊗N ′ (see p. 27), so tensor descends toa function t : F × F → F taking

[M ], [N ]

7→ [M ⊗N ], which in turn restricts to a function t1 : F1× F1 → F1. By(2.14.i,ii,iv), t and t1 are commutative, associative binary operations on F , F1 with identity element [A], and so makeF a monoid with submonoid F1. Writing Z[F ] and Z[F1] for the monoid rings ([5.33]), the bijections C ↔ Z[F ]and C1 ↔ Z[F1] (hereafter taken as identifications) define ring structures on these groups; write τ and τ1 for themultiplications.

If M ∈ F1 and 0 → N ′ → N → N ′′ → 0 is a a short exact sequence of A-modules, then by flatness of Mthis induces a short exact sequence 0 → M ⊗N ′ → M ⊗N → M ⊗N ′′ → 0, so τ1

[M ], [N ′]− [N ] + [N ′′]

=[M ⊗N ′]− [M ⊗N ]+ [N ⊗N ′′] ∈D1 for any generator [N ′]− [N ]+ [N ′′] of D1, showing D1 is an ideal, and thusgiving a ring structure on K1 =C1/D1.24 Now γ1(M ) · γ1(N ) = (π1 τ1)(M , N ) =π1

[M ⊗N ]

= γ1(M ⊗N ). Since

22 If 0→ N → M → P → 0 is an exact sequence of finitely generated A-modules and N and P are flat, so is M by [2.25]. This isn’t strictlynecessary for the definition, which only requires those sequences such that all terms are flat, but it’s somehow comforting anyway.

23 It is tempting, in order to do i) and ii), to consider K1 as a subset of K and argue the desired multiplication on K1 is a restriction of the mapµ : K1 ×K → K making K a K1-module. In general, it is not; the map ε : K1 → K defined in the remark by ε

γ1(M )

= γ (M ) is not in generalinjective. The problem is that, while C1 ⊆ C naturally, we don’t necessarily have D1 = C1 ∩D . Indeed, D1 is generated by [N ]− [M ]+ [P ] forshort exact sequences 0→N →M → P → 0 of objects inF1, but it is entirely possible that there are elements of C1 ∩D not generated by thesesequences.

24 It’s worth noting that K generally fails to be a ring precisely because D is not generally an ideal, which is in turn the case because modulesare not flat in general.

106


[A] is the neutral element of F1 (which shows A is flat, hence inF1), it becomes a unity in the monoid ring C1, andits image γ1(A) ∈K1 is the unity of K1.

ii) Show that tensor product induces a K1(A)-module structure on the group K(A), such that γ1(M ) · γ (N ) = γ (M ⊗N ).Note that the (restricted) multiplication τ : C1×C →C makes the group C a C1-module. A slight modification

of the proof in the preceding paragraph that D1 ÃC1 is an ideal, assuming N ′, N , N ′′ ∈F and not necessarily inF1,shows that τ(C1×D)⊆D , so K =C/D is naturally a C1-module. If D1 ⊆AnnC1

(K), then (p. 19) we may naturallyregard K as a K1-module. Indeed, for any N ∈F and short exact sequence 0→ M ′→ M → M ′′→ 0 inF1 we havea Tor exact sequence ([2.24]) Tor1(M

′′, N )→ M ′⊗N → M ⊗N → M ′′⊗N → 0, and Tor1(M′′, N ) = 0 by flatness

of M ′′ ([2.24]), so τ

[M ′]− [M ]+ [M ′′], [N ]

∈ D ; since these [M ′]− [M ]+ [M ′′] are generators for D1, it followsD1 ·K = 0. Now γ1(M ) · γ (N ) = (π τ)(M , N ) =π

[M ⊗N ]

= γ (M ⊗N ).

iii) If A is a (Noetherian) local ring, then K1(A)∼=Z.Note that the proof of [7.26.i] transfers verbatim to show that K1 satisfies an analogous universal property:

namely for any abelian group G, there is a bijective correspondence between additive functions λ : F1 → G andgroup homomorphisms λ0 : K1→G given by λ= λ0 γ1. Then as in [7.26.iii], the additive function rkA :F1→ Z,with range N, induces a surjective homomorphism K1→Z, and so K1 contains an infinite cyclic additive subgroup.Recall from [7.15] that if A is Noetherian and local, then M ∈ F1 if and only if M ∼= An for some n ∈ N. Since theγ1(M ) generate K1, it follows the additive group of K1 is cyclic; hence K1

∼=Z.

iv) Let f : A → B be a ring homomorphism, B being Noetherian. Show that extension of scalars gives rise to a ringhomomorphism f ! : K1(A)→K1(B) such that f !

γ1(M )

= γ1(B⊗AM ). If g : B→C is another ring homomorphism(with C Noetherian), then (g f )! = g ! f !.(2.20) states that for M a flat A-module, M | f := B⊗AM is a flat B -module, and (2.17) that if M is finitely generated,

so is M | f . Thus we have a map | f :F1(A)→F1(B). Set λ = γ1,B | f :F1(A)→F1(B)→ K1(B). If 0→ N → M →P → 0 is a short exact sequence in F1(A), then the proof of parts i,ii) shows that 0→ N | f → M | f → P | f → 0 isexact, so

N | f

−

M | f

+

P | f

∈D1(B), showing λ is additive. By the universal property in the proof of part iii), itfollows there is a unique group homomorphism f! : K1(A)→K1(B) such that f!γ1,A= γ1,B | f . Further, this is a ringhomomorphism, for B⊗AA∼= B by (2.14.iv), and (B⊗AM )⊗B (B⊗AN )∼=M ⊗AB⊗B B⊗AN ∼=M ⊗A(B⊗B B)⊗AN ∼=M ⊗AB ⊗AN ∼= B ⊗A(M ⊗AN ) by (2.14.i,ii,iv) and (2.15) (viewing B as a (B , A)-bimodule).

For M ∈ F1(A) we have

M | f

g = C ⊗B M | f = C ⊗B (B ⊗AM ) ∼= (C ⊗B B)⊗AM ∼= C ⊗AM = M |g f by (2.15) (viewing B as a (B , A)-bimodule) and(2.14.iv). If we write α f for the map F1(A) → F1(B) induced by | f , and so on,then we’ve shown αg f = αg α f : F1(A) → F1(C ). Now (g f )! is the uniquehomomorphism ν : K1(A)→K1(C ) such that νγ1,A= γ1,C |g f , and so is uniquesuch that ν π1,A = π1,C αg f . On the other hand, the diagram at right showsν = g ! f ! also satisfies this equality, so (g f )! = g ! f !

F1(A)

| f// F1(B)

|g// F1(C )

F1(A)

π1,A

α f

// F1(B)

π1,B

αg

// F1(C )

π1,C

K1(A) f !

// K1(B) g !// K1(C ).

v) If f : A→ B is a finite ring homomorphism then

f!

f !(x)y

= x f!(y)

for x ∈K1(A), y ∈K(B). In other words, regarding K(B) as a K1(A)-module by restriction of scalars, the homomorphismf ! is a K1(A)-module homomorphism

By bi-additivity of the module multiplication, since f ! and f! are homomorphisms, and since elements x =γ1,A(M ) and y = γB (N ) generate K1(A) and K(B), it will suffice to check the equality for these elements:

f!

f !γ1,A(M )

· γB (N )

iv)= f!

γ1,B (B ⊗AM ) · γB (N ) ii)= f!

γB

(M ⊗AB)⊗B N

(2.15)=

(2.14.iv)f!

γB (M ⊗AN ) [7.26.iv]= γA

(M ⊗AN )| f

= γA(M ⊗AN | f )ii)= γ1,A(M ) · γA(N | f )

[7.26]= γ1,A(M ) · f!

γB (N )

.

107


It doesn’t make sense to consider f ! to be a K1(A)-module homomorphism K(A) → K(B); we defined f ! asa map K1(A) → K1(B) because a short exact sequence in F (A) has no reason to remain exact under B ⊗A− forarbitrary finite A-algebras B . So assume instead that the book actually meant to ask about the homomorphismf! : K(B) → K(A).25 One can define a map t ′ : F1(A)×F (B) → F (B) by t ′(M , N ) = M ⊗AN | f , first restrictingscalars along f and then regarding the result as a B -module by b (m ⊗ n) = m ⊗ b n. This then induces a bilinearmap τ′ : C1(A)×C (B)→ C (B) making C (B) a C1(A)-module. Thinking of B -modules as A-modules by restrictionof scalars, our proof of part ii) above gives an induced map µ′ : K1(A)×K(B)→K(B)making K(B) a K1(A)-module.We claim µ′(x, y) =µB

f !(x), y); it again suffices to check for generators x = γ1,A(M ) and y = γB (N ). Now

µB

f !(x), y) =µB

γ1,B (B ⊗AM ), γB (N )

= γB

(B ⊗AM )⊗B N

= γB (M ⊗AN ) =µ′(x, y).

Remark. Since F1(A) is a subset of F (A)we have a group homomorphism ε : K1(A)→K(A) given by ε

γ1(M )

= γ (M ).If the ring A is finite-dimensional and regular, i.e., if all its local rings Ap are regular (Chapter 11) it can be shownthat ε is an isomorphism.

25 To regard K(B) as a K1(A)-module by restriction of scalars, one’s first inclination is to set x · y = x · f!(y), but the former is supposed to bein K(B) and the latter is in K(A); apparently this is not what the book means by “restriction of scalars.”

108

Chapter 8: Artin Rings

Example. (p. 91, top) If A = k[x1, x2, . . .] is a polynomial ring in countably many indeterminates over a field k,the ideal a is (x1, x2

2, x33, . . .), and B = A/a, then writing yn = xn , the book claims that m = (y1, y2, . . .) is the only

prime ideal of B . Evidently m is maximal, since k[y1, y2, . . .]/(y1, y2, . . .) ∼= k. Since each yn ∈ N(B) is nilpotent,we have m ⊆ N(B), so by (1.8) every prime of B contains m, which is then minimal, hence the only prime. But(y1)( (y1, y2)( · · · is an infinite ascending sequence of ideals, so B is not Noetherian. Three other examples can befound in the solution to [7.8].

EXERCISES1. Let q1 ∩ · · · ∩ qn = 0 be a minimal primary decomposition of the zero ideal in a Noetherian ring [A], and let qi be

pi -primary. Let p(m)i be the mth symbolic power of pi (Chapter 4, Exercise 13). Show that for each i = 1, . . . , n there

exists an integer mi such that p(mi )i ⊆ qi .

Set q= qi , p= pi , and Sp =A\p. From (7.14), since the radical r (q) = p, it follows from (7.14) that there is m ∈Nsuch that pm ⊆ q. Then p(m) = Sp(p

m)⊆ Sp(q) = q by (4.12*.iv,iii).

Suppose qi is an isolated primary component. Then Apiis an Artin local ring, hence if mi is its maximal ideal we will

have mri = 0 for all sufficiently large r, hence qi = p(r )i

Since p(r ) ⊆ q for sufficiently large r, and p(r ) is p-primary by [7.13.i], we have another primary decompositionp(r ) ∩

⋂

j 6=i q j = 0. By the uniqueness (4.11) of isolated primary components, it follows q= p(r ).1

Now we claim Ap is a local Artinian ring with nilpotent maximal ideal m := pe . By (3.13), Ap is local withmaximal ideal m, and by (7.3) it is Noetherian. By (4.6), p is minimal in V (0) = Spec(A); it follows from (3.13) thatm is also minimal in Ap, hence the only prime, so dim(Ap) = 0 and Ap is Artinian by (8.5). Further, m=N(Ap) by(1.8), so by (7.15) or (8.4) there is r ∈N such that mr = 0.2

If qi is an embedded primary component, then Apiis not Artinian, hence the powers mr

i are all distinct, and so thep(r )i are all distinct. Hence in the given primary decomposition we can replace qi by any of the infinite set of pi -primary

ideals p(r )i where r ≥ ri , and so there are infinitely many minimal primary decompositions of 0 which differ only in thepi component.

If we have p j ( p with q j another primary component, then pej ( pe in Ai , so dim(Ai )≥ 1 and Ai is not Artinian

by (8.5). It follows from (8.6) (using contraposition to exclude case ii)) that all the powersmr are distinct. If pr = pr+1

for some r, then taking extensions and using (3.11.v) we have mr = (pe )r = (pr )e = (pr+1)e = (pe )r+1 = mr+1, soit follows the pr are all distinct. Since A is Noetherian, pr has a primary decomposition by (7.13); by our proof of[4.13.ii], p(r ) is the smallest primary ideal containing pr . It follows that since the pr are distinct, so are the p(r ). Nowif r ≥ ri , we have 0 6= p(r ) ( p(ri ), so any of these p(r ) can be substituted in the primary decomposition of (0).

1 Despite the utter triviality of this two-line argument, I had to poach it fromhttp://scribd.com/doc/47338424/atiyah-macdonald-solutions.

2 I was unable to prove the “hence” of the problem; while I’ve proven the nilpotence of m and that q= p(r ) separately, I still don’t know whythe latter follows from the former.

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http://scribd.com/doc/47338424/atiyah-macdonald-solutions

Ex. 8.2 Chapter 8: Artin Rings

2. Let A be a Noetherian ring. Prove that the following are equivalent:i) A is Artinian;

ii) Spec(A) is discrete and finite;iii) Spec(A) is discrete.

ii) =⇒ iii): Trivial.iii) =⇒ i): If Spec(A) is discrete, then in particular each point is closed. By [1.18.i], the closed points correspond

to maximal ideals, so every prime in A is maximal. Then dim(A) = 0, so by (8.5), A is Artinian.i) =⇒ ii): If A is Artinian, every prime ideal of A is maximal by (8.1), so by [1.18.i] or [3.11], each point of

X = Spec(A) is closed. By (8.3) it only has a finite number of maximal (hence prime ideals), so it is finite. But thenevery subset of X is a finite union of closed sets, hence closed, and X is discrete.

3. Let k be a field and A a finitely generated k-algebra. Prove that the following are equivalent:i) A is Artinian;

ii) A is a finite k-algebra.i) =⇒ ii): By (8.7), A is a finite direct product of local Artinian rings Aj , which are quotients of A under the

projection, and hence again finitely generated (by the images of the generators of A, for instance). Thus if we canprove each of the finitely many Aj is a finite k-algebra, so will A be.

So without loss of generality assume A is local Artinian ring finitely generated over k, with lone prime m. ThenB =A/m is a field, again finitely generated over k, so by Zariski’s Lemma ((1.27.2*), (5.24), [5.18], (7.9)), B is a finitealgebraic extension of k, and hence a finite k-vector space. Since primary decompositions exist in the Noetherianring A and Spec(A) = m, we see m belongs to (0). Then as A is finitely generated as an A-module, [7.18] givesus a chain 0 = M0 ( M1 ( · · · ( Mn = A of A-submodules whose successive quotients are of the form A/pi forpi ∈ Spec(A) = m — that is to say, all isomorphic to B . Thus we have a finite collection of short exact sequences0 → Mi → Mi+1 → B → 0 of A-modules, which we may view as k-modules. Since dimk is an additive function,this gives us dimk Mi+1 = dimk Mi + dimk B , and taking i = 0 shows dimk M1 = dimk B is finite. By induction,dimk A= n dimk B is finite.

ii) =⇒ i): A finite k-algebra (p. 30) is finitely generated as a k-module. By (6.10), it follows A satisfies the d.c.c.on k-submodules. Since each A-module is naturally also a k-module, it follows A satisfies the d.c.c. on ideals as well,and so is an Artinian ring.

4. Let f : A→ B be a ring homomorphism of finite type. Consider the following statements:i) f is finite;

ii) the fibres of f ∗ are discrete subspaces of Spec(B);iii) for each prime ideal p of A, the ring B ⊗Ak(p) is a finite k(p)-algebra ( k(p) is the residue field of Ap);iv) the fibres of f ∗ are finite.

Prove that i) =⇒ ii) ⇐⇒ iii) =⇒ iv).If f is integral and the fibres of f ∗ are finite, is f necessarily finite?For the last question, it is important to realize we are not assuming finite type; otherwise, (5.2) and the Remark

on p. 60 show the answer is yes. Consider the algebraic closure K of a field K , constructed for example in [1.13].3

The fibers of the inclusion K ,→K will be finite, for both spectra are one-point spaces. However, this extension willnot be finite unless dimK K is finite, which is not the case, for example, if K is a finite algebraic extension of anymember of Fp ,Q,Qp , k(S), where k is any field and S 6=∅ any set of indeterminates.

For the implications, fix p ∈ Spec(A) and let k = k(p) =Ap/pAp be the residue field of the localization Ap. Recallfrom (2.14.i) that C := B ⊗Ak ∼= k ⊗AB and from [3.21.iv] that Spec(C ) ≈ ( f ∗)−1(p). Note further, using (2.14.iv)and (2.15), that

C := B ⊗Ak ∼= B ⊗A(Ap⊗Apk)∼= (B ⊗AAp)⊗Ap

k . (8.1)

i) =⇒ iii): Assume B is generated as an A-module by n elements. It follows from the proof of (2.17) that B⊗AAp

is generated by n elements as an Ap-module and by (2.17) again and Eq. 8.1 that C is a ≤ n-dimensional k-vectorspace. Since C is a k-algebra, it is then a finite k-algebra.

iii) =⇒ ii): By the assumption and [8.3], C is an Artinian ring. Then by [8.2], its spectrum ( f ∗)−1(p) is discrete.

3 We defined K as the subset of elements of L =⋃∞

n=0 Kn algebraic over K = K0, where each Kn+1 is the smallest algebraic extension of Kn

in which each irreducible monic polynomial p(x) ∈ Kn[x] has a root. But it turns out K1 = K = L. By (5.4) and induction, each Kn is integralover K , and each α ∈ K ⊆ L is in some Kn , so K is integral over K . Since, each Kn is integral over K , each member thereof satisfies a polynomialequation p(x) ∈K[x]; but that shows that p(x) already has a root in K1, which then must be itself algebraically closed.

110

Chapter 8: Artin Rings Ex. 8.5

ii) =⇒ iii): We first show C is a finitely generated k-algebra (and hence, by (7.7), Noetherian).4 Since f is of finitetype, there is a surjective ring homomorphism A[x1, . . . , xn]→ B . Recalling from (2.18) that tensor is right exact,we apply Ap ⊗A− to both sides and use [2.6] (see the footnote to [7.13]) to get a surjective ring homomorphismAp[x1, . . . , xn] → Ap ⊗A B . Applying k ⊗Ap

− to both sides, the same argument, followed by (2.15), (2.14.i), and

Eq. 8.1, yields a surjective ring homomorphism k[x1, . . . , xn]→ k ⊗Ap

Ap ⊗AB ∼= C , so C is a finitely generated

k-algebra. By (7.7), C is a Noetherian ring, and by assumption, ( f ∗)−1(p)≈ Spec(C ) is discrete, so [8.2] shows C isArtinian. Because of this and because C is a finitely generated k-algebra, [8.3] shows C is a finite k-algebra.

ii) & iii) =⇒ iv): By iii), C is a finite k-algebra, hence a fortiori finitely generated, and hence by (7.7) a Noetherianring; and by ii), ( f ∗)−1(p)≈ Spec(C ) is discrete; so from [7.2] it follows Spec(C ) = ( f ∗)−1(p) is also finite.

5. In Chapter 5, Exercise 16, show that X is a finite covering of L (i.e., the number of points of X lying over a given point ofL is finite and bounded.)

Refer back to our solution to [5.16] for notation. In summary, we were given an affine subvariety X ⊆ kn andfrom it constructed a linear surjection π : kn k r = L such that π|X is already surjective. The coordinate rings([1.27]) of X and L were written, respectively, as A and A′, and r and π were chosen in such a way that the injection$ = (π|X )# : A′ A was integral of finite type. View it as an inclusion. Recalling from [1.27] and [5.16] that wemay identify X with Max(A) and the map X → L with its induced map Max(A)→Max(A′), we now want to showonly finitely many maximal ideals of A lie over any maximal ideal of A′. Using (5.2) again, $ is finite, so by [8.4] itfollows the fibers of$∗ are finite.

However, we still want to uniformly bound the size of these fibers. Since $ is integral of finite type, by theRemark on p. 60, $ is finite; say A is generated as an A′-module by n elements. Then n is a uniform bound on thek-dimension of B = A⊗A′k(p) for p ∈ Spec(A′), and by [3.21.iv] it is enough to show Spec(B) has ≤ n points. Letp1, . . . , pm be distinct primes of B . Since they are coprime, the canonical map B→

∏

B/pi is surjective. Since B is ak(p)-algebra, the pi and hence the B/pi are k(p)-vector spaces as well. We have n ≥ dimk B ≥

∑mi=1 dimk (B/pi )≥ m,

so Spec(B) is finite.5

6. Let A be a Noetherian ring and q a p-primary ideal in A. Consider chains of primary ideals from q to p. Show that allsuch chains are of finite bounded length, and that all maximal chains have the same length.

We have a bijection, by (1.1) and (3.9), between the set Σ = a Ã A : q ⊆ a ⊆ p and the ideals (a/q)p/q ofB = (A/q)p/q, which we claim preserves and reflects being (p-)primary. By (4.8), extension along A/q→ B preservesbeing primary (for a ⊆ p/q), and by p. 50, contraction along A→ B preserves being primary, so it remains to seeextension a 7→ a/q along A A/q does, for a⊇ q. But for x, y ∈ A, x y ∈ a/q implies xy ∈ a, so that x ∈ a or someyn ∈ q, meaning x ∈ a/q or yn = yn ∈ a/q.

Note that as a localization of a quotient of a Noetherian ring, B is also Noetherian, by (7.1) and (7.4). Since A isNoetherian and r (q) = p, (7.14) gives an n ≥ 1 such that pn ⊆ q, so the maximal ideal m of the local ring B satisfiesmn = 0. By (8.6), B is Artinian. Also, for any bÃ B we have mn ⊆ b⊆m, so by (7.16), b is m-primary.6

Thus our question boils down to arbitrary chains of ideals in an Artinian ring B . By the definition of the word,these must all have finite length. A maximal chain is a composition series (p. 76), and by (6.7), these all have the samelength.

4 It feels like this should have been proven in the book somewhere already, but I don’t see where, so I’m doing it here.5 A slightly different proof, from [Milne, Prop. 8.5], is as follows. Let m′ ∈Max(A′) and K = A′/m′; then by (1.17.i), each m ∈Max(A) with

mc =m′ has (m′)e =mc e ⊆m, so m descends to a maximal ideal of A/(m′)e . Since A is a finite A′-module, with n generators for some n, by (2.8)C = A/(m′)e is a ≤ n-dimensional K -algebra. If Spec(C ) contains m primes, then the same proof as above, with C replacing B , shows n ≥ m,and this limits the number of maximal m lying over m′.

6 Amusingly, this seems to show the ideals of Σ were all p-primary. Can that be right?

111

Chapter 9: Discrete Valuation Rings and DedekindDomains

EXERCISES1. Let A be a Dedekind domain, S a multiplicatively closed subset of A. Show that S−1A is either a Dedekind domain or the

field of fractions of A.By (9.3), A is an integrally closed Noetherian domain of dimension one. (5.12) implies that S−1A is also integrally

closed and (7.3) that it is Noetherian. S−1A is a domain since it is contained in the field of fractions K of A. By(3.11.iv), the longest possible chain of prime ideals in S−1A is (0) = S−1(0)⊆ S−1p for a prime pÃA not meeting S.Since p 6= 0 and S−1 are contained in a field, S−1p 6= (0). It follows from the definition (9.3) that S−1A is a Dedekinddomain if there is some prime p disjoint from S. If there is none, then (0) is maximal in S−1A, which is then a field.Since it contains A, it then is K .

Suppose that S 6= A\0, and let H , H ′ be the ideal class groups of A and S−1A respectively. Show that extension ofideals induces a surjective homomorphism H →H ′.

First we show a 7→ S−1a is a homomorphism of fractional ideal groups I (A)→ I (S−1A). Since A is Noetherian,each fractional ideal a of A is finitely generated; it follows from (3.15) then that S−1(A : a) = (S−1A : S−1a), soS−1a−1 = (S−1a)−1. For multiplication, S−1S−1 = S−1 since 1 ∈ S, so S−1(ab) = S−1S−1ab= (S−1a)(S−1b).

This map is surjective because a 7→ S−1a is surjective on integral ideals (3.11.i) and fractional ideals of a Noethe-rian ring are of the form x−1a for x ∈A and aÃA (p. 96). For x ∈K , we have xA 7→ S−1(xA) = x(S−1A), so principalfractional ideals are mapped to principal fractional ideals, and the surjective homomorphism I (A) → I (S−1A) de-scends to a surjective homomorphism H →H ′.

2. Let A be a Dedekind domain. If f = a0+ a1x + · · ·+ an xn is a polynomial with coefficients in A, the content of f is theideal c( f ) = (a0, . . . , an) in A.

Prove Gauss’s lemma that c( f g ) = c( f )c(g ).Let f =

∑

ai x i , g =∑

b j x j , and f g =∑

ck xk , where ck =∑

i+ j=k ai b j . We will have c( f g ) ⊆ c( f )c(g ) just iffor each prime pÃA, c( f g )p =

c( f )c(g )

p.1 But

c( f )c(g )

p= c( f )pc(g )p by (3.11.v), and c( f )p is the content of

the image of f in Ap. Thus, by the definition (9.3), we may assume A is a discrete valuation ring (henceforth “DVR”).Finitely generated ideals of A are principal2, so we can write c( f ) = (a′) and c(g ) = (b ′) and we know a′b ′ divides

f g in A[x]. Note that if d is a divisor of each coefficient of h ∈A[x], so that h/d ∈A[x], we have c(h) = d · c(h/d ).Thus (a′) = c( f ) = a′ · c( f/a′), so c( f/a′) = (1), and similarly c(g/b ′) = (1). In the terminology of [1.2.iv], f/a′ andg/b ′ are primitive, by the result of that exercise, the same holds of f g/a′b ′, which then has content (1). It followsthat c( f )c(g ) = (a′b ′) = a′b ′ · c( f g/a′b ′) = c( f g ).

1 Let N , P ⊆ M be A- modules. To show N = P it is enough to show that the inclusions N ,→N + P and P ,→N + P are surjective. By (3.9)and (3.4.i), this happens if and only if for all primes p Ã A, Np ,→ (N + P )p = Np + Pp and Pp ,→ Np + Pp are surjective; but this happens justwhen Np = Pp.

2 Such a ring is called a Bézout domain because it satisfies Bézout’s lemma that if d is a common divisor of a, b ∈A, there exist y, z ∈A such thatd = ay+b z; see http://en.wikipedia.org/wiki/Bézout_domain and http://planetmath.org/encyclopedia/BezoutDomain.html.

A ring satisfying Gauß’s lemma is called a Gaussian ring, and the Gaussian rings that are domains turn out (see e.g. http://arxiv.org/abs/1107.0440) to be exactly the Prüfer domains (see http://en.wikipedia.org/wiki/Prüfer_domain). These rings have many charac-terizations, one of which is that all their localizations at primes are valuation rings, others being that the nonzero finitely generated ideals are allinvertible and that all their ideals are flat. Cf. also http://planetmath.org/encyclopedia/PruferRing.html.

If one instead defines the content of a polynomial as the greatest common divisor of its coefficients (which will generate the ideal we calledthe content before, in the event this ideal is principal) one can extend the result to integral domains such that any two elements have greatestcommon divisors, and in particular to unique factorization domains. This is easily proved, e.g., at http://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomial)#A_proof_valid_over_any_GCD_domain. See also http://planetmath.org/GcdDomain.html and http://planetmath.org/encyclopedia/PropertiesOfAGcdDomain.html.

112

http://en.wikipedia.org/wiki/Bzout_domain

http://planetmath.org/encyclopedia/BezoutDomain.html

http://arxiv.org/abs/1107.0440


http://en.wikipedia.org/wiki/Prfer_domain

http://planetmath.org/encyclopedia/PruferRing.html

http://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomial)#A_proof_valid_over_any_GCD_domain

http://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomial)#A_proof_valid_over_any_GCD_domain

http://planetmath.org/GcdDomain.html

http://planetmath.org/encyclopedia/PropertiesOfAGcdDomain.html

http://planetmath.org/encyclopedia/PropertiesOfAGcdDomain.html

Chapter 9: Discrete Valuation Rings and Dedekind Domains Ex. 9.3

3. A valuation ring (other than a field) is Noetherian if and only if it is a discrete valuation ring.The argument on p. 94 shows a DVR is a Noetherian valuation ring (and not a field).For the other direction, by (5.18), a Noetherian valuation ring A is local. It is an integral domain by the definition

on p. 65. It now will be enough, by (9.7), to show the non-zero fractional ideals of A are invertible, or by (9.2), toshow the maximal ideal m is principal and A has dimension one. First we show A is a PID. Any ideal a of A is finitelygenerated as (x1, . . . , xn) for some x j ∈A since A is Noetherian. By [5.28], the ideals (x j ) are totally ordered, so onecontains all the others, and hence a is principal.

To use (9.7), just note that since A is a Noetherian PID, any fractional ideal of A is of of the form x−1(y) for somex, y ∈A (p. 96) and so has an inverse y−1(x).

To use (9.2), set m= (m). Any prime ideal (p) satisfies (p)⊆ (m), so for some a ∈A we have am = p and hence,since (p) is prime, either m ∈ (p) or a ∈ (p). If the former holds, p=m. If the latter holds, we have b ∈A such thatp = am = b pm, or (1− bm)p = 0. As m is not a unit 1− bm 6= 0, so since A is an integral domain, p = 0. It followsthat the only chain of prime ideals in A is (0)(m, meaning A has dimension one.

4. Let A be a local domain which is not a field and in which the maximal ideal m is principal and⋂∞

n=1mn = 0. Prove that

A is a discrete valuation ring.Let p be a generator for m, so for each n ≥ 0 we have mn = (pn). Since

⋂∞n=1m

n = 0, it follows there is no nwith mn = mn+1, and every x ∈ A\0 fails to be in some mn , so there is a greatest number v(x) ∈ N such thatpv(x) divides x (taking p0 = 1). If v(x) = n, we can write x = upn for some u ∈ A; since pn+16 | x, it follows thatu ∈ A\m= A×. Thus A\0 ∼= A×× pN as a multiplicative monoid. It follows that for every pair x, y ∈ A we havev(x)≤ v(y) ⇐⇒ x|y. Therefore any ideal a is generated by an element x ∈ a with v(x)minimal, so that the idealsof A are (0) and the (pn). This shows A is Noetherian of dimension one. By any of the implications iii), v), vi) =⇒i) of (9.2), A is a DVR.

5. Let M be a finitely-generated module over a Dedekind domain. Prove that M is flat ⇐⇒ M is torsion-free.Let the Dedekind domain be A. Since A is an integral domain, by [3.13], M is torsion-free just if for each prime

p Ã A we have Mp torsion-free over the DVR Ap. Now each Mp is finitely generated over a PID, and so3 can bewritten as the direct sum of the torsion submodule and a free module. Thus each Mp is torsion-free if and only if itis free. Since A is Noetherian and M is finitely generated, by [7.16], M is flat just if each of the Mp is free. It followsthat M is flat if and only if it is torsion-free.

6. Let M be a finitely generated torsion module (T (M ) = M) over a Dedekind domain A. Prove that M is uniquely repre-sentable as a finite direct sum of modules A/pni

i , where pi are nonzero prime ideals of A.For each prime ideal p of A we have Ap a DVR. Since Mp is finitely generated and torsion, and Ap a PID (p. 94),

the structure theorem for finitely generated modules shows4 Mp is isomorphic to a direct sum of modules Ap/(d j )for d j ∈Ap, where the (d j ) are primary and unique up to order. But since Ap is a DVR, each (d j ) = (pAp)

n j for somen j ≥ 1. So Mp

∼=⊕

j Ap/pn j Ap. By exactness of localization (3.3) and [1.21.iv], since p is the only prime ideal of A

containing pn j we have Ap/pn j Ap

∼= (A/pn j )p ∼=A/pn j , so each Mp is of the form we desire for M . It will now sufficeto show M is isomorphic to the direct sum of finitely many Mp.

Since M is finitely generated, [3.19.v] shows Supp(M ) = V

Ann(M )

. As A is a Dedekind domain, the prime-power factorization of Ann(M ) shows that Supp(M ) is finite. Now the canonical maps m 7→ m/1: M →Mp for eachp ∈ Supp(M ) naturally compile into a map φ : M →

⊕

Mp. Note that since each pair p 6= q of primes is coprime,there are x ∈ pn \q for arbitrarily high n, which then annihilate the (finitely generated) summands A/pn of Mp,so by [3.1], (Mp)q = 0 for distinct primes p and q. On the other hand, (Mp)p ∼= Mp.5 Therefore, since localizationdistributes over direct sums (Eq. 3.6) localizing φ at q shows φq : Mq→

⊕

Mp

q∼= (Mq)q ∼=Mq is an isomorphism

for q ∈ Supp(M ); and similarly φq : Mq→⊕

Mp

q∼=⊕

0 is an isomorphism 0→ 0 for q /∈ Supp(M ). By (3.9), φ isan isomorphism.

3 http://planetmath.org/encyclopedia/FinitelyGeneratedModulesOverAPrincipalIdealDomain.html4 See http://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_

domain#Primary_decomposition.5 For a multiplicative submonoid S of A, the map a 7→ (a/1)/1: A→ S−1(S−1A) satisfies the conditions (3.2), so that S−1A ∼= S−1(S−1A),

Using (3.5), we can rewrite this as S−1A ∼= S−1A⊗AS−1A, so for any A-module M , using (3.5) again and (2.14.ii), S−1M ∼= S−1A⊗AM ∼=S−1A⊗AS−1A⊗AM ∼= S−1(S−1M ).

113

http://planetmath.org/encyclopedia/FinitelyGeneratedModulesOverAPrincipalIdealDomain.html

http://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain#Primary_decomposition

http://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain#Primary_decomposition

Ex. 9.7 Chapter 9: Discrete Valuation Rings and Dedekind Domains

7. Let A be a Dedekind domain and a 6= 0 an ideal in A. Show that every ideal in A/a is principal.First, let p be a prime and n ≥ 1. As in the previous exercise, A/pn ∼=Ap/p

nAp, and Ap is a DVR, hence (p. 94) aPID, so each ideal of A/pn is principal.

Given any a Ã A, use (9.1) to produce a prime factorization a =∏

pnii and consider the standard map A→

∏

A/pnii . Since the pni

i are coprime, (1.10) says that a=⋂

pnii and the map is surjective with kernel a. Thus we can

write A/a∼=∏

A/pnii as a product of rings Bi each of whose ideals is principal. As in [1.22], each ideal b of A/a is a

sum of ideals of the Bi . If ei is the element of A/awhose i -component is 1 and whose other components are zero, andb has bei = (bi ei ) for bi ∈ Bi , then b=

∑

bei is generated by the single element b = ⟨bi ⟩ ∈A/a whose i -componentis bi , since b ∈ b and each bi ei = b ei ∈ (b ). Thus A/a is principal.

Deduce that every ideal in A can be generated by at most 2 elements.Suppose c Ã A is not principal, and let a nonzero a ∈ c be given. Then the ideal c/(a) of A/(a) is principal,

generated by b +(a) for some b ∈ c. It follows that c= (a, b ) in A.6

8. Let a, b, c be three ideals in a Dedekind domain. Prove that

a∩ (b+ c) = (a∩ b)+ (a∩ c),a+(b∩ c) = (a+ b)∩ (a+ c).

Writing I for the set of nonzero ideals of our Dedekind domain, the problem is asking us to prove the lattice(I , +, ∩) is distributive. Since all the ideals in question are submodules of A, by the first footnote to [9.2], to provean equation it suffices to show the localizations of the sides at each prime agree. By (3.4), localization distributesover sum and intersection, so we now only have to prove the equations for A a DVR. But pm + pn = pminm, n andpm ∩ pn = pmaxm, n, so pn 7→ n is a lattice isomorphism (I , +, ∩)→ (N, min, max). Since the latter is distributive,7

the equations hold, and we are done.8

6 A converse also holds: if A is a domain such that for every aÃA and nonzero a ∈ a there exists b such that (a, b ) = a, then A is a Dedekinddomain.

To see this, note that A is Noetherian, and all its localizations Ap at primes pmust also be Noetherian by (7.3) and satisfy the same two-generatorproperty. Let 0 6= aÃAp. Picking a nonzero element a ∈ma⊆ a and applying the two-generator property to a, we see there must be b ∈ a suchthat a= (a, b ) =ma+(b ) in Ap. Now a is finitely generated and m=R(Ap), so by the corollary (2.7) to Nakayama’s Lemma, a= (b ). Thus Ap

is a local, Noetherian PID, and so by (9.2) is a DVR. Since this holds for all p, by (9.3) A is a Dedekind domain.This result can apparently be attributed to a C.-H. Sah; see Theorem 20.11 of Pete L. Clark’s notes http://math.uga.edu/~pete/integral.

pdf.7 To be thorough about this, we show that the lattice given by a totally ordered set (like N), when equipped with the operations x ∨ y =

maxx, y and x ∧ y =minx, y, satisfies the equations x ∨ (y ∧ z) = (x ∨ y)∧ (x ∨ z) and x ∧ (y ∨ z) = (x ∧ y)∨ (x ∧ z) for all x, y, z. Since ∨and ∧ are symmetric in their arguments, we may assume without loss of generality that y ≤ z. To prove the equations, using a bit more bruteforce than we would like, we tabulate the values of the relevant terms and check that the sides are equal given any of the three possible orderingsof x, y, z.

x ∨ (y ∧ z) (x ∨ y) ∧ (x ∨ z) x ∧ (y ∨ z) (x ∧ y) ∨ (x ∧ z)x y x z

x ≤ y ≤ z y z x xy y x x

y ≤ x ≤ z x z y xx x x x

y ≤ z ≤ x x x y zx x z z

8 It is not strictly necessary to localize. For A a Dedekind domain again, the prime factorization (9.1) shows that to prove an equation itwill be enough to show vp-values of the sides agree for all primes p, where vp is given by a =

∏

p∈P pvp(a). To do this, we show vp is a lattice

homomorphism (I , +, ∩)→ (N, min, max). Since A has dimension one, its primes are maximal and hence pairwise coprime. This allows us toconclude products of disjoint sets of prime ideals are coprime, as follows: if ideals ai and b j are such that ai + b j = (1) for all i and j , then∏

ai +∏

b j = (1), for (1) =∏

i (ai + b j ) ⊆∏

ai

+ b j , so (1) =∏

j

∏

i

ai

+ b j

⊆∏

i ai +∏

j b j . Thus by (1.10.i), each intersection

of powers of distinct primes is actually a product, and vice versa. Products of ideals do distribute over sums (p. 6), and for m ≤ n we haveam + an = am and am ∩ an = an . Write np =minvp(a), vp(b) and Np =maxvp(a), vp(b). Then

a+ b=∏

pnp∏

pvp(a)−np +∏

pnp∏

pvp(b)−np =∏

pnp ·∏

pvp(a)−np +∏

pvp(b)−np

=∏

pnp

since the two terms in the parentheses share no prime factors in common. Similarly,

a∩ b=∏

pvp(a) ∩∏

pvp(b) =⋂

pvp(a) ∩⋂

pvp(b) =⋂

pvp(a) ∩ pvp(b)

=⋂

pNp =∏

pNp .

We now have vp(a+ b) =minvp(a), vp(b) and vp(a ∩ b) =maxvp(a), vp(b) as hoped.

114



Chapter 9: Discrete Valuation Rings and Dedekind Domains Ex. 9.9

9. (Chinese Remainder Theorem). Let a1, . . . , an be ideals and let x1, . . . , xn be elements in a Dedekind domain A. Then thesystem of congruences x ≡ xi (mod ai ) (1≤ i ≤ n) has a solution x in A ⇐⇒ xi ≡ x j (mod ai + a j ) whenever i 6= j .

Following the book’s hint, define φ : A→⊕n

i=1 A/ai by φ(x)i = x+ai , and ψ :⊕n

i=1 A/ai →⊕

i< j A/(ai +a j )byψ

⟨xi+ai ⟩

⟨i , j ⟩ = xi− x j +ai+a j . The system of congruences x ≡ xi (mod ai ) has a solution x just if ⟨xi+ai ⟩ ∈im(φ), and the conditions xi ≡ x j (mod ai+a j ) are satisfied just if ⟨xi+ai ⟩ ∈ ker(ψ). Then the statement in questionis true just if the sequence

Aφ→⊕

A/aiψ→⊕

A/(ai + a j )

is exact. But this means just to show im(φ) = ker(ψ), and thus by the first footnote to [9.3] it is enough to show itis true after localizing at each prime p. By (3.4), localization distributes over quotients and sums, so it is enough toprove the results for ideals ai = pki of a DVR A. Without loss of generality, assume ki ≤ k j for i < j .

If ⟨xi+pki ⟩ ∈ ker(ψ), it follows that for i < j we have xi− x j ∈ pki +pk j = pki . Thenφ(xn) = ⟨xi+p

ki ⟩, showingker(ψ)⊆ im(φ). That ψ φ= 0 holds in any ring:

(ψ φ)(x)

⟨i , j ⟩ =ψ

⟨x + ai ⟩

⟨i , j ⟩ = x − x + ai + a j = ai + a j .

115

Chapter 10: Completions

More than any other chapter, this one leaves small, eminently believable statements unproved. Before tackling theexercises, we prove some of these assertions. Doing so takes a surprisingly long time.

Lemma 10.1. Let H be the intersection of all neighborhoods of 0 in [a topological abelian group] G. Theni) H is a subgroup.

The book notes that “i) follows from the continuity of the group operations.” This is true, as far as it goes, butis actually longer, in its details, than the rest of the proof.

Write NG(0) for the set of neighborhoods of 0 (N (0) when G is understood). Note that since x 7→ −x is ahomeomorphism, for each neighborhood U ∈ N (0) we also have −U ∈ N (0). Then V = U ∩−U ∈ N (0) andV = −V . If x ∈ H and U ∈ N (0), find a subset V ∈ N (0) with V = −V . Then x ∈ V = −V , so −x ∈ V ⊆ U .Since U was arbitrary, −x ∈H .

Since + : G×G→G is continuous and sends ⟨0, 0⟩ 7→ 0, for any U ∈N (0) there is a neighborhood W of ⟨0, 0⟩in G×G that addition maps into U . By the definition of the product topology, then, there are V1, V2 ∈N (0) suchthat V1 ×V2 ⊆W . If we set V = V1 ∩V2, then V +V ⊆ U . Now suppose x, y ∈ H , and let U ∈ N (0). There isV ∈N (0) such that V +V ⊆U , and x, y ∈V , so x + y ∈U . As U was arbitrary, x + y ∈H .

Finally, note 0 ∈H , so H is nonempty.

Equivalence of Cauchy sequences is an equivalence relation.* (p. 102)⟨xν⟩ is equivalent to ⟨xν⟩ since the differences xν − xν are identically zero.If ⟨xν⟩ is equivalent to ⟨yν⟩, then by definition xν−yν → 0, which we take to mean that for every U ∈N (0) there

is t (U ) ∈N such that for all ν ≥ t (U )we have xν−yν ∈U . To show that ⟨yν⟩ is also equivalent to ⟨xν⟩, let U ∈N (U )be given and find a subset V ∈N (0) with V =−V . For ν ≥ t (V ) we have xν − yν ∈V =−V , so yν − xν ∈V ⊆U .

Now suppose ⟨xν⟩ is equivalent to ⟨yν⟩ and ⟨yν⟩ is equivalent to ⟨zν⟩. To show ⟨xν⟩ is equivalent to ⟨yν⟩, let U ∈N (0) be given, and let V ∈N (0) be such that V +V ⊆U . By assumption, there are numbers t (V ), t ′(V ) ∈N suchthat for ν ≥ t (V ) we have xν − yν ∈ V and for ν ≥ t ′(V ) we have yν − zν ∈ V . For ν ≥ maxt (V ), t ′(V ) we havexν − zν = (xν − yν )+ (yν − zν ) ∈V +V ⊆U .

If ⟨xν⟩, ⟨yν⟩ are Cauchy sequences, so is ⟨xν + yν⟩, and its class in G depends only on the classes of ⟨xν⟩ and ⟨yν⟩.Let U ∈ N (0), and let V ∈ N (0) be such that V +V ⊆ U . Since ⟨xν⟩ and ⟨yν⟩ are Cauchy, there are numbers

s(V ) and s ′(V ) such that xµ − xν ∈ V for all µ, ν ≥ s(V ) and yµ − yν ∈ V for all µ, ν ≥ s ′(V ). Then for allµ, ν ≥maxs(V ), s ′(V ) we have (xµ+ yµ)− (xν + yν ) = (xµ− xν ) + (yµ− yν ) ∈V +V ⊆ U . As U was arbitrary,⟨xν + yν⟩ is Cauchy.

If ⟨x ′ν⟩ represents the same class as ⟨xν⟩, so that xν − x ′ν → 0, then ⟨xν + yν⟩ and ⟨x ′ν + yν⟩ are equivalent, since(xν + yν )− (x ′ν + yν ) = (xν − x ′ν ) + (yν − yν ) = xν − x ′ν → 0. Similarly, if ⟨y ′ν⟩ represents the same class as ⟨yν⟩, then⟨x ′ν + yν⟩ and ⟨x ′ν + y ′ν⟩ are equivalent, so by transitivity ⟨xν + yν⟩ and ⟨x ′ν + y ′ν⟩ are equivalent.

Similarly, if ⟨xν⟩ is a Cauchy sequence, then ⟨−xν⟩ is a Cauchy sequence whose class in G depends only on that of ⟨xν⟩.*This and the next are very easy, but necessary to show G is a group. Let U ∈N (0) be given; since ⟨xν⟩ is Cauchy,

there is s(U ) ∈N such that µ, ν ≥ s(U ) implies xµ− xν ∈ U . But (−xν )− (−xµ) =−(xν − xµ) = xµ− xν ∈ U then,for all ν, µ≥ s(U ). As U was arbitrary, ⟨−xν⟩ is Cauchy.

If ⟨x ′ν⟩ represents the same class as ⟨xν⟩, so that xν−x ′ν → 0, then (−xν )−(−x ′ν )→ 0 as well; for given any U ∈N (0)we may find a smaller V = −V ∈ N (0) and some t (V ) such that for all ν ≥ t (V ) we have xν − x ′ν ∈ V = −V , so(−xν )− (−x ′ν ) =−(xν − x ′ν ) ∈V ⊆U as well.

116


⟨0⟩ is a Cauchy sequence.*This is trivial, since all differences are zero.

Hence we have an addition in G with respect to which G is an abelian group.By the last two facts on independence of representatives, it will be enough to show the set of Cauchy sequences

in G forms an Abelian group. But this easily follows from the facts that this group is defined as a subgroup of theproduct group GN and that the abelian group identities hold componentwise:

⟨xν⟩+ ⟨yν⟩

+ ⟨zν⟩= ⟨xν + yν⟩+ ⟨zν⟩=

(xν + yν )+ zν

=

xν +(yν + zν )

= ⟨xν⟩+ ⟨yν + zν⟩= ⟨xν⟩+

⟨yν⟩+ ⟨zν⟩

;

⟨xν⟩+ ⟨0⟩= ⟨xν + 0⟩= ⟨xν⟩;

⟨xν⟩+ ⟨−xν⟩=

xν +(−xν )

= ⟨0⟩;

⟨xν⟩+ ⟨yν⟩= ⟨xν + yν⟩= ⟨yν + xν⟩= ⟨yν⟩+ ⟨xν⟩.

For each x ∈ G the class of the constant sequence ⟨x⟩ is an element φ(x) of G, and φ : G → G is a homomorphism ofabelian groups. [The kernel of φ is the subgroup

⋂

N (0) of (10.1).]This may be too obvious to bother with, but we do it anyway. Constant sequences are Cauchy because the

differences x − x = 0 of ⟨x⟩ approach (are) 0. φ is obviously a homomorphism: φ(0) is the equivalence class of ⟨0⟩,which is the zero of G; ⟨−x⟩=−⟨x⟩, so taking classes, φ(−x) = −φ(x); and ⟨x + y⟩= ⟨x⟩+ ⟨y⟩, so taking classes,φ(x + y) = φ(x) +φ(y). We have x ∈ ker(φ) just if ⟨x⟩ is in the class of ⟨0⟩, so that x = x − 0→ 0 as the indicesincrease. But this means that for each U ∈ N (0) there is t (U ) such that for ν ≥ t (U ) we have x ∈ U . Since x isindependent of ν, that just means x ∈U , so x ∈

⋂

N (0).

There is a natural topology making G a topological group.*

For each U ∈NG(0), define U ⊆ G to be the set of all elements x ∈ G such that all representatives ⟨xν⟩ of x are“eventually in” U :

U :=¦

x ∈ G : ∀⟨xν⟩ ∈ x ∃N ∈N ∀ν ≥N (xν ∈U )©

.1

Note that for all U , V ∈NG(0) we have 0 ∈ØU ∩V = U ∩ V . Then if we write let N = U : U ∈NG(0) and takeall translates in G, these sets together generate a unique topology on G for which N is a neighborhood basis of 0.2

Suppose x, y ∈ G are such that their sum lies in an open set W ⊆ G. By our definition of the topology, there isa U ∈ N such that (x+ y)+ U ⊆W . By our proof of (10.1.i), there is V ∈NG(0) such that V +V ⊆U . If ⟨xν⟩ ⟨yν⟩are Cauchy sequences representing elements of V , then there is N ∈ N sufficiently large that for all ν ≥N we havexν , yν ∈ V , and hence xν + yν ∈ U . Thus V + V ⊆ U , so addition takes

x + V

×

y + V

into (x + y) + U ⊆Wand hence is continuous.

Similarly, any open set about −x, contains a basic open set −x + U . By our proof of (10.1.i) again, there is aV =−V ∈NG(0) such that V ⊆ U . Then the opposite of any sequence eventually in V is also eventually in V , so−V = V , and −

x + V

=−x + V ⊆−x + U , so inversion is also continuous.

φ : G→ G is continuous.*

Let x ∈G and consider a basic open neighborhood φ(x)+ U . If u ∈U ∈NG(0), there is some V ∈NG(0) suchthat u+V ⊆U . Thus, if ⟨yν⟩ is a Cauchy sequence equivalent to ⟨u⟩, we must eventually have yν− u ∈V , so that yνis eventually in u +V ⊆U . This shows φ(u) ∈ U . It follows that φ(x +U )⊆φ(x)+ U , showing φ is continuous.

1 We do have to stipulate eventual containment for all sequences in a class: in ⟨Q, +⟩, for instance, the Cauchy sequences ⟨ n−1n ⟩ and ⟨1⟩ are

equivalent, but the former is always in the open ball of radius 1 about 0, while the latter never is.It took me a while to decide between this definition and several other less fruitful possibilities, and I eventually set-

tled on this one as a result of Gerald A. Edgar’s answer at http://math.stackexchange.com/questions/192808/topology-induced-by-the-completion-of-a-topological-group.

2 Brian M. Scott’s answer at http://math.stackexchange.com/questions/67259/inquiry-regarding-neighborhood-baseselaborates how this works. But there is a question as to how to prove that in the topology generated by thetranslates of N , these sets are actually neighborhoods: http://math.stackexchange.com/questions/234803/translating-a-neighborhood-basis-of-a-topological-group.

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http://math.stackexchange.com/questions/192808/topology-induced-by-the-completion-of-a-topological-group

http://math.stackexchange.com/questions/192808/topology-induced-by-the-completion-of-a-topological-group

http://math.stackexchange.com/questions/67259/inquiry-regarding-neighborhood-bases

http://math.stackexchange.com/questions/234803/translating-a-neighborhood-basis-of-a-topological-group

http://math.stackexchange.com/questions/234803/translating-a-neighborhood-basis-of-a-topological-group


If G is first countable or Hausdorff, then G is first countable or Hausdorff, respectively.*

If ⟨Ui ⟩i∈N is a countable neighborhood basis of 0 in G, we claim ⟨Ui ⟩ will be a countable neighborhood basis of0 in G. Indeed, for any V ∈ NG(0) there is Ui ⊆ V , so for any V ∈ N there is Ui ⊆ V . By translation, we havecountable neighborhood bases at every point of G.

Given two distinct points of G, ginding disjoint neighborhoods of the two is equivalent, by translation, to find-ing disjoint neighborhoods of 0 and their difference x. Let ⟨0⟩ and ⟨xν⟩ be representatives. Since they are assumedinequivalent, it follows there is some U ∈NG(0) such that the differences xν are not eventually in U , so that x /∈ U .Using the proof of (10.1.i), find V ∈NG(0) such that V +V ⊆U . Now we claim the basic neighborhoods V 3 0 andx + V 3 x are disjoint. If not, there would be y in their intersection, so that y ∈ V and y ∈ x + V , or equivalentlyy− x ∈ V . But then from our proof G is a topological group we would have x = y+(y− x) ∈ V +V ⊆ U , after all.

If G is first countable and Hausdorff, then G is complete in the sense that all Cauchy sequences in G converge to points ofG.*3

Let ⟨Ui ⟩i∈N be a countable neighborhood basis of 0 in G. By the proof of (10.1.i), we may assume Ui =−Ui andUi+1+Ui+1 ⊆Ui for all n, so that the same will hold of ⟨Ui ⟩.

Let ⟨xn⟩ be a Cauchy sequence in G, and select a representative ⟨xn, ν⟩ for each xn . For each n, since the sequence⟨xn, ν⟩ is Cauchy, there is a number Λn such that when ν , µ ≥ Λn , we have xn, ν − xn,µ ∈ Un+2. Since the sequence⟨xn⟩ is Cauchy, there is for each i ∈N an Ni ∈N such that for all n, m ≥ Ni we have xn − xm ∈ Ui+2. That in turnimplies we have xn, ν − xm, ν ∈Ui+2 for all sufficiently large ν.

Now for each ν ∈N, let yν = xν,Λν ∈G. We claim that ⟨yν⟩ is a Cauchy sequence. Indeed, if n, m ≥Ni , there arearbitrarily large ν such that xn, ν − xm, ν ∈Ui+2, in particular such that ν ≥maxΛn , Λm, and then

yn − ym = xn,Λn− xm,Λm

= (xn,Λn− xn, ν )

︸︷︷︸

ν ≥ Λn

+ (xn, ν − xm, ν ) + (xm, ν − xm,Λm)

︸︷︷︸

ν ≥ Λm

∈Ui+2+Ui+2+Ui+2 ⊆Ui . (10.1)

Let y ∈ G be the equivalence class of ⟨yν⟩. Our goal is to show xn → y, or in other words that for each Ui ,we have y − xn ∈ Ui for all sufficiently large n. In other words, we want to show that if n is big enough then eachrepresentative ⟨zn, ν⟩ of y − xn is in Ui for all high enough ν. It will actually be enough to find Mi such that foreach n ≥ Mi , the representative ⟨wn, ν⟩= ⟨yν − xn, ν⟩ is eventually in Ui+1; for then, given n ≥ Mi and any otherrepresentative ⟨zn, ν⟩, we will have zn, ν −wn, ν eventually in Ui+1, so that zn, ν = (zn, ν −wn, ν ) +wn, ν is eventually inUi+1+Ui+1 ⊆Ui .

Mi =Ni+2 from above will work, for if m ≥Ni+1 and ν ≥ Ξm :=maxΛm , Ni+1, then we indeed have

yν − xm, ν = (yν − ym)+ (ym − xm, ν ) = (yν − ym)︸︷︷︸

Eq. 10.1: m, ν ≥ Ni+1

+ (xm,Λm− xm, ν )

︸︷︷︸

ν ≥ Ξm ≥ Λm

∈ Ui+1+Ui+2 ⊆Ui .

If H is another abelian group and f : G→H a continuous homomorphism, then the image under f of a Cauchy sequencein G is a Cauchy sequence in H , and therefore f induces a homomorphism f : G→ H, which is continuous.

Let ⟨xν⟩ be a Cauchy sequence in G and let U ∈ NH (0) be given. Since f (0) = 0, by continuity, there is V ∈NG(0) such that f (V ) ⊆ U . By assumption there is s(V ) ∈ N such that xµ − xν ∈ V for all µ, ν ≥ s(V ). Thenf (xµ)− f (xν ) = f (xµ− xν ) ∈ f (V )⊆U . Thus

f (xν )

is Cauchy.

To show f is well defined, we must show it preserves equivalence. But if ⟨xν⟩ and ⟨x ′ν⟩ are equivalent, then thereis t (V ) ∈ N such that xν − x ′ν ∈ V for ν ≥ t (V ), so f (xν )− f (x ′ν ) = f (xν − x ′ν ) ∈ f (V ) ⊆ U , showing

f (xν )

and

f (x ′ν )

are equivalent.

It is obvious f is a homomorphism because operations on Cauchy sequence are defined componentwise and fis induced by applying f componentwise to Cauchy sequences.

3 This is also a special case of a general result in Bourbaki’s General Topology Part 1 (Chapter III, §3.5, Theorem I) regarding completions (interms of Cauchy filters) with respect to the right uniformity, but I have not attempted to translate that proof into our language.

Our case could also be proved using the theorem that a first-countable, Hausdorff topological group is metrizable; see http://u.math.biu.ac.il/~megereli/mickey25.pdf.

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http://u.math.biu.ac.il/~megereli/mickey25.pdf

http://u.math.biu.ac.il/~megereli/mickey25.pdf


To prove continuity we use the topology on the completion we defined above. Let a basic neighborhood f (x)+Vin H be given, where V ∈NH (0) Since f is continuous and f (0) = 0, there is U ∈NG(0) such that f (U )⊆V . Thenfor each Cauchy sequence ⟨uν⟩ eventually in U , the image sequence ⟨ f (uν )⟩ is eventually in V , so that f (U )⊆ V andtherefore f (x + U )⊆ f (x)+ V , showing f is continuous.

Thus we have a sequence of subgroupsG =G0 ⊇G1 ⊇ · · · ⊇Gn ⊇ · · ·

and U ⊆G is a neighborhood of 0 if and only if it contains some Gn ....

In fact if g ∈ Gn then g +Gn is a neighborhood of g ; since g +Gn ⊆ Gn this shows Gn is open. [Prove that taking thex +Gn as a neighborhood basis of x defines a topology on G making G a topological group.*] (pp. 102–3)

The fact that the authors prove the Gn are open shows they are using the Bourbaki definition of neighborhoods:4

to wit, given a topology on X , a set N ⊆ X is defined to be a neighborhood of x just if there is an open set U ⊆ Xwith x ∈ U ⊆N . A fundamental system of neighborhoods (neighborhood basis) of a point x ∈ X is then a collectionB of subsets of X such that the neighborhoods of x are precisely the sets containing a member ofB . On the otherhand, given collectionsB(x) of subsets of X , one for each x ∈ X , the following axioms guarantee that they definea unique topology under which eachB(x) is a neighborhood basis for x:

1. if N , N ′ ∈B(x), there there is N ′′ ∈B(x) with N ′′ ⊆N ∩N ′;

2. x is in each member ofB(x);

3. if N ∈B(x), there is N ′ ∈B(x) such that for all y ∈N ′ there exists N ′′ ∈B(y) with N ′′ ⊆N

(“any neighborhood of x is also a neighborhood of all points sufficiently near x”).

It is not hard to see that that the set of translates x+Gn satisfies these axioms: 1. for m ≤ n, evidently x+Gn ⊆(x +Gm)∩ (x +Gn); 2. obviously x ∈ x +Gn ; 3. for all y ∈ x +Gn we have y +Gn = x +Gn +Gn = x +Gn . Thuswe have a well-defined topology on G. Inversion is continuous, since it sends x +Gn ↔−x +Gn , and addition iscontinuous since, given a basic neighborhood x+y+Gn , addition sends the neighborhood (x+Gn)×(y+Gn)⊆G×Gof ⟨x, y⟩ into it.

If 0→G′i,→G

p→G′′→ 0 is an exact sequence of groups and Gn ⊆G is a subgroup, then

0→ G′

G′ ∩Gn

ι→ GGn

π→ G′′

p(Gn)→ 0

is an exact sequence.* (pp. 104–5)The kernel of G′ ,→GG/Gn is G′∩Gn , so ι is defined and injective. The composition G

pG′′G′′/p(Gn)

is surjective and takes the same value on any member of a class g+Gn , so π is defined and surjective. Since p i = 0,so also π ι = 0. Now suppose g +Gn ∈ ker(π). Then p(g ) ∈ p(Gn), so there is h ∈ Gn with p(g − h) = 0. Byexactness of the original sequence, g−h ∈G′. Now ι

g−h+(G′∩Gn)

= g+Gn , so the sequence is exact at G/Gn .

[W]e can apply (10.3) with G′ =Gn[;] then G′′ =G/Gn has the discrete topology so that G′′ =G′′.From p. 103 recall that Gn is open, so its translates g +Gn are open. Since these become points in G′′, it follows

G′′ is discrete. Using the neighborhood 0 of 0 in G′′, it follows that for every Cauchy sequence ⟨xν⟩ in G′′, there issome s(0) such that for allµ, ν ≥ s(0)we have xµ− xν = 0; that is, every Cauchy sequence is eventually constant.

But a sequence that is eventually x is equivalent to the constant sequence ⟨x⟩, so the canonical map G′′→ G′′ is anisomorphism.

Since the an are ideals it is not hard to check that with [the a-topology]A is a topological ring, i.e., that the ring operationsare continuous.

4 Nicolas Bourbaki, General Topology Part 1, Definitions 4 and 5, pp. 18–21.

119


We have shown above that A is a topological group under addition; it remains to show multiplication is contin-uous. But multiplication maps the product neighborhood (x+an)× (y+an) of ⟨x, y⟩ into xy+ xan + yan +a2n ⊆xy + an .

The completion A of A is again a topological ring[.]In the light of (10.4), the unspecified topology in question should be given by letting the x+Òan be a neighborhood

basis for each x ∈ A. Because ⟨Òan⟩ is a decreasing sequence of subgroups, A a topological group under addition. EachÒan is an ideal, since it is the contraction of the ideal

∏∞j=1 a

n/(a j ∩an) in the product ring∏∞

j=1 A/a j . Thus, as with

A, multiplication A×A→ A takes the neighborhood (x+Òan)×(y+Òan) of ⟨x, y⟩ into xy+xÒan+yÒan+Óa2n ⊆ xy+Òan ,so that A is a topological ring.

φ : A→ A is a continuous ring homomorphism, whose kernel is⋂

an .By p. 102, φ is a group homomorphism with kernel as stated; multiplicativity follows because ⟨x⟩⟨y⟩ = ⟨xy⟩,

and taking classes gives φ(x)φ(y) = φ(xy). As for continuity, for x ∈ an we have φ(x) represented in∏∞

j=1 A/a j

by ⟨x + a j ⟩ ∈∏∞

j=1 an/(a j ∩ an), so that φ(an) ⊆ Òan ; thus for any y ∈ A and any basic neighborhood φ(y) + Òan of

φ(y) ∈ A we have φ(y + an)⊆φ(y)+ Òan .

Likewise for an A-module M : take G = M , Gn = an M . This defines the a-topology on M , [making M a continuousA-module.]

Because Gn is a decreasing sequence of additive subgroups, this does indeed define a topology on M . To seecontinuity, note that given a basic neighborhood x m + an M of x m ∈ M , the map A× M → M takes the openneighborhood (x+an)× (m+an M ) of ⟨x, m⟩ into (x+an)(m+an M ) = x m+an m+ xan M +a2n M ⊆ x m+an M .

[T]he completion M of M is a topological A-module.The topology on M is defined, as before by the basic neighborhoods Õan M of 0. Here we note that the module

multiplication A× M → M is left undefined. For elements ⟨x j + a j ⟩ ∈ A and ⟨m j + a j M ⟩ ∈ M , define their productas ⟨x j +a j ⟩ · ⟨m j +a j M ⟩= ⟨x j m j +a j M ⟩. To see this works, first note that x j+1m j+1− x j m j = x j+1(m j+1−m j )+

(x j+1 − x j )m j ∈ x j+1aj+1M + a j+1m j ⊆ a j+1M , so the resulting sequence is in M . Second, if we replaced x j with

x ′j ∈ x j + a j , then we would have (x ′j − x j )m j ∈ a j M , and similarly if we replaced m j with m′j ∈ m j + a j M , so thatthis multiplication is well defined.

Now, observe ÒakÕan M ⊆Õak+n M . Indeed, if x j ∈ ak and m j ∈ an M for all j , then x j m j ∈ ak+n M for all j , so that

the sequence elements x j m j + a j M are in ak+n M/(a j M + ak+n M ). For continuity, taking x ∈ A and m ∈ M , and

considering the basic neighborhood x m+Õan M of their product, note that multiplication takes (x+Òan)×(m+Õan M )to x m+ Òan m+ xÕan M + ÒanÕan M ⊆ x m+Õan M .

If f : M →N is any A-module homomorphism, [... f] defines f : M → N .Let f

⟨m j+aj M ⟩

= ⟨ f (m j )+aj N ⟩. This is an element of N since f (m j+1)− f (m j ) = f (m j+1−m j ) ∈ f (a j+1M )⊆

a j+1N . To see the map is well defined, note that if m′j −m j ∈ a j M , then f (m′j ) ∈ f (m j )+ a j N .

f is a group homomorphism because it is defined in terms of the homomorphisms f : M/a j M →N/a j N givenby f (m + a j M ) = f (m) + a j N . It is a module homomorphism because f

⟨x j m j + a j M ⟩

= ⟨x j f (m j ) + a j N ⟩ =

⟨x j + a j ⟩ f

⟨m j + a j M ⟩

whenever ⟨x j + a j ⟩ ∈ A and ⟨m j + a j M ⟩ ∈ M .

f is also continuous, because f

Õan M

⊆ÕAnN , as f takes sequences of elements of an M/(a j M+an M ) to sequencesof elements of anN/(a j N + anN ).

If C is a ring, and A=C [x1, . . . , xn] a polynomial ring, and a= (x1, . . . , xn) the ideal of polynomials with no constantterm, then A=C [[x1, . . . , xn]], the ring of formal power series.*

This generalizes Example 1) on this page.am ÃA is the set of polynomials with no terms of degree < m. If we let b= (x1, . . . , xn) in B :=C [[x1, . . . , xn]],

then bm is the set of power series of order≥ m. Since every class modulo bm is represented by the (polynomial) class

120


of its truncation below degree m, we have isomorphisms B/bm ∼=A/am compatible with the maps m+1 7→ m, andso B ∼= A. But the natural map B→ B is an isomorphism: it is injective since

⋂

bm = 0, and surjective because if pm

is the sum of terms of bm+1 of total degree n and ⟨bm + bm⟩ ∈ B , then∑

p j + bm =∑

j<m pm + bm = bm + bm foreach m.

[A]ll stable a-filtrations on M determine the same topology on M , namely the a-topology. (p. 106)The proof of (10.6) shows any stable a-filtrations ⟨Mn⟩ and ⟨M ′n⟩ have bounded difference, so we may fix n0 such

that for all n ∈ N we have Mn+n0⊆ M ′n and M ′n+n0

⊆ Mn . Let x ∈ M . For any ⟨Mn⟩-basic neighborhood x +Mn

each point y ∈ x +Mn has a ⟨M ′n⟩-basic neighborhood y +M ′n+n0⊆ y +Mn = x +Mn , so x +Mn is open in the

⟨M ′n⟩-topology, and thus the ⟨M ′n⟩-topology contains the ⟨Mn⟩-topology. The converse holds by symmetry. Thus allstable a-filtrations on M induce the same topology the stable filtration ⟨an M ⟩ does, namely the a-topology.

Let A be a ring (not graded), a an ideal of A. Then we can form a graded ring A∗ =⊕∞

n=0 an . Similarly, if M is an

A-module and Mn is an a-filtration of M , then M ∗ =⊕

n Mn is a graded A∗-module, since am Mn ⊆Mm+n . (p. 107)We can afford some clarification of what is going on here. Writing An = an , we have A∗ =

⊕

An , and ourmultiplication takes Am×An→Am+n . The slightly confusing thing is that there also is an am ⊆A=A0, for instance,and if we multiply that by An , we get a subset of am+n ⊆An . So what the book is saying is that if we multiply Am bythe nth summand Mn of M ∗, we get a subset of the (m+ n)th summand Mm+n ; we are not considering, say, Mn ⊆Min the zeroth summand of M ∗.

Proposition 10.3 12 *. Using (10.3) or otherwise it is clear that a-adic completion commutes with finite direct sums. (p.

108)By induction it will suffice to show a-adic completion distributes over binary direct sums. It is possible to use

(10.3) and the diagrammatic definition of direct sums to do this, but easier to just give an explicit isomorphism. Thea-adic filtration on M⊕N is ⟨an M⊕anN ⟩. It is clear

⟨xn+an M ⟩, ⟨yn+a

nN ⟩

7→

⟨xn , yn⟩+an(M⊕N )

gives a well-

defined, homomorphic bijectionφ : M⊕N ↔ÙM ⊕N since we have natural isomorphisms (M/an M )⊕(N/anN ) ∼−→(M⊕N )/an(M⊕N ) and ⟨xn+1, yn+1⟩ ≡ ⟨xn , yn⟩

modan(M⊕N )

if and only if xn+1 ≡ xn (mod an M ) and yn+1 ≡ yn(mod anN ).

[I]f F ∼=An we have A⊗AF ∼= F .

A⊗AF ∼= A⊗AAn = A⊗A

n⊕

j=1

A(2.14.iii)∼=

n⊕

j=1

(A⊗AA)(2.14.iv)∼=

n⊕

j=1

A(10.3 1

2 *)∼=

×n⊕

j=1

A= F

A⊗AN //

A⊗AF

N // F .*

Given an A-module homomorphism φ : N → F , the following square commutes:The map around the upper-right is the composition

A⊗AN → A⊗AF → A⊗AF → A⊗AF → F taking

⟨an + an⟩⊗ x 7→ ⟨an + an⟩⊗φ(x) 7→ ⟨an + an⟩⊗

φ(x)+ an F

7→ ⟨an + an⟩⊗

φ(x)+ an F

7→

anφ(x)+ an F

,

and the map around the lower-left is the composition A⊗AN → A⊗AN → A⊗AN → N → F given by

⟨an + an⟩⊗ x 7→ ⟨an + an⟩⊗ ⟨x + anN ⟩ 7→ ⟨an + an⟩⊗ ⟨x + anN ⟩ 7→ ⟨an x + anN ⟩ 7→

φ(an x)+ an F

.

[In the following diagram, if the rows are exact, γ is surjective, and β is injective, then a] little diagram chasing provesthat α is injective:

Nζ //

γ

Fη //

β

M //

α

0

N ε// F

δ// M .

121

Ex. 10.1 Chapter 10: Completions

Suppose m ∈ ker(α). There is f ∈ η−1(m), and δβ f = αη f = 0. Since the bottom row is exact, there is n ∈ Nwith εn = β f , and since γ is surjective, there is n ∈ γ−1(n). Now εγ n = βζ n = β f , so by injectivity of β we getf = ζ n, and thus m = η f = ηζ n = 0, showing α is injective.

[C]heck that xm xn does not depend on the particular representatives chosen. [Here we have a ring A with ideal a and aregiven xm ∈ am , xn ∈ an . We write xm ∈ am/am+1, xn ∈ an/an+1, xm xn := xm xn ∈ am+n/am+n+1]* (p. 111)

As sets we have (xm + am+1)(xn + an+1) = xm xn + xnam+1+ xma

n+1+ am+n+2 ⊆ xm xn + am+n+1.

Similarly, if M is an A-module and ⟨Mn⟩ is an a-filtration of M , . . .

G(M ) =∞⊕

n=0

Mn/Mn+1

. . . is a graded G(A)-module in a natural way.Let ak ∈ ak and xn ∈Mn ; since ⟨Mn⟩ is an a-filtration, ak xn ∈Mn+k . Note that

(ak + ak+1)(xn +Mn+1) = ak xn + ak+1xn + ak Mn+1+ ak+1Mn+1 ⊆ ak xn +Mn+k+1,

so for ak ∈ ak/ak+1 and xn ∈Mn/Mn+1 we can define ak xn := ak xn ∈Mn+k/Mn+k+1 uniquely. Extend this definitionby distributivity to a product G(A)×G(M ) → G(M ), so that to finish checking G(M ) is a G(A)-module, it onlyremains to check 1x = x and (ab )x = a(b x) for a, b ∈ A and for x ∈ G(M ). Using distributivity, we only need tocheck for homogeneous elements; but then it is obvious, for 1xn = 1xn = xn and

(ak b`)xn = ak b` xn = ak b`xn = ak b`xn = ak (b` xn) in Mk+`+n/Mk+`+n+1

simply because the associative rule holds for the A-module M . To see G(M ) is a graded G(A)-module, note that byconstruction Gk (A)Gn(M ) = (a

k/ak+1)(Mn/Mn+1)⊆Mn+k/Mn+k+1 =Gn+k (M ).

If β φ= φ α with β : M → M injective, α a bijection, and φ surjective, then φ is surjective.* (p. 113)Write ψ = φ α; it is surjective since α and φ are. Thus M = im(ψ) = β

im(φ)

⊆ β(M ); so β is surjective aswell, hence a bijection. Thus surjectivity of φ follows from that of ψ.

EXERCISES1. Let αn : Z/pZ→Z/pnZ be the injection of Abelian groups given by αn(1) = pn−1 and let α : A→ B be the direct sum of

all the αn (where A is a countable direct sum of copies ofZ/pZ, and B is the direct sum of theZ/pnZ). Show that the p-adiccompletion of A is just A but that the completion of A for the topology induced from the p-adic topology on B is the directproduct of the Z/pZ. Deduce that the p-adic completion is not a right-exact functor on the category of all Z-modules.

Since pA = 0, the sequence A/pnA is · · ·A id→ Aid→ A, the coherent sequences of which are constant sequences

⟨a⟩, giving an obvious isomorphism lim←− A/pnA∼=A.To make coordinate references easier, let Gn be the nth copy of Z/pZ in A. We have pnB =

⊕

j>n pnZ/p jZ.As im(α j ) = p j−1Z/p jZ, which is contained in pnZ/p jZ so long as j > n, it follows α−1

j (pnZ/p jZ) is Z/pZ if

j > n and 0 otherwise. Then An := α−1(pnB) is the subgroup⊕

j>n G j of A=⊕∞

j=1 G j , so A/An∼=⊕n

j=1 G j . Theprojection A/An+1→A/An kills the (n+1)st component and preserves the others; the inverse system associated to thetopology on A induced byα is thus essentially · · · → (Z/pZ)3→ (Z/pZ)2→Z/pZ. Writingπ j : A/An+1→G j for theprojection, in a coherent sequence ⟨ξn⟩, the component πn(ξn) ∈ Gn determines (is equal to) the the componentsπn(ξ j ) of all the later members ξ j , j ≥ n, so the map φ : lim←− A/An →

∏∞n=1 Gn taking ⟨ξn⟩ 7→ ⟨πn(ξn)⟩ is an

isomorphism.Consider the short exact sequence 0 → A

α→ B → B/α(A) → 0. If we topologize the groups by the filtrations⟨α−1(pnB)⟩, ⟨pnB⟩, and ⟨pnB/α(A)⟩, then by (10.3) the corresponding sequence of completed systems is exact; soif we instead give A the p-adic (discrete) topology ⟨pnA= 0⟩, then, assuming the map A→ B remains defined, theresulting sequence should not still be exact. Indeed, we have maps A/pnA∼= A→ α(A) ,→ B → B/pnB compiling

into an inverse system, so we have a short sequence 0→A→ B→ÚB/α(A)→ 0. Since A 6∼=∏

nZ/pZ, this sequence isnot exact at B , so p-adic completion is not right-exact. (Though it does preserve surjectivity, because if ρ : BC is a

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Chapter 10: Completions Ex. 10.2

surjection, we have ρ(pnB) = pnC , so we have a surjective map ⟨B/pnB⟩→ ⟨C/pnC ⟩ of surjective inverse systems,and (10.3) gives us B C .)5 p-adic completion is not left-exact, by the same example; the essential reason is thatgiven a general homomorphism α : A→ B (in our example, an injection), we needn’t have α(pnA) = α(A)∩ pnB , sothe p-adic topology on A is not that induced from B and the hypotheses of (10.3) are not met.

2. In Exercise 1, let An = α−1(pnB), and consider the exact sequence

0→An→A→A/An→ 0.

Show that lim←− is not right exact, and compute lim←−1 An .

We can rewrite the sequence as 0 →⊕∞

j=n+1 G j →⊕∞

j=1 G j →⊕n

j=1 G j → 0. The inclusions An+1 ,→ An ,identity map idA, and projections A/An+1 A/An give us an exact sequence of inverse systems. (10.2) gives us anexact sequence

0→ lim←− An→ lim←− A→ lim←− A/An→ lim←−1 An→ lim←−

1 A→ lim←−1 A/An→ 0.

To show lim←− is not right exact, it will be enough, by this sequence, to show lim←−1 An 6= 0. Since ⟨A⟩ and ⟨A/An⟩

are surjective systems, the proof of (10.2) shows us lim←−1 A and lim←−

1 A/An are 0. As in the last problem, lim←− A = Aand lim←− A/An

∼=∏∞

j=1 G j . For lim←− An , since the maps are inclusions, any coherent sequence is a constant sequence,which means its lone term must be in each An . But

⋂∞n=1 An = 0. Thus our exact sequence actually gives us a short

exact sequence

0→Aψ→ lim←− A/An→ lim←−

1 An→ 0.

To identify the last term it remains to describe the injection ψ. We claim it is basically the inclusion⊕∞

j=1 G j ,→∏∞

j=1 G j . Indeed, since the maps A→ A/An are quotient maps, the map from A∼= lim←− A(∏∞n=1 A to lim←− A/An (

∏∞n=1 A/An is given by a 7→ ⟨a⟩ 7→ ⟨a (mod An)⟩. Composing with the isomorphisms A/An

∼=⊕n

j=1 G j , this givesa 7→

⟨π j (a)⟩nj=1

∞n=1, sending a ∈

∏∞j=1 G j to a list of truncations. Using our isomorphism φ : ⟨ξn⟩ 7→ ⟨πn(ξn)⟩

from the last problem finally gives a 7→ ⟨πn(a)⟩= a. Thus lim←−1 An∼=∏∞

n=1Z/pZ⊕∞

n=1Z/pZ

.

3. Let A be a Noetherian ring, a an ideal and M a finitely-generated A-module. Using Krull’s Theorem and Exercise 14 ofChapter 3, prove that

∞⋂

n=1

an M =⋂

m⊇aker(M →Mm),

where m runs over all maximal ideals containing a.Krull’s Theorem (10.17) says that the left-hand side is the set of elements of M annihilated by 1+a for some a ∈ a,

so it remains to show the same is true of the right. Let x ∈M . For all maximal m⊇ a we have 1+a⊆ 1+m⊆A\mso if (1+ a)x = 0 for some a ∈ a, then by [3.1], x/1= 0 in Mm for all maximal m⊇ a. On the other hand if x/1= 0in Mm for all maximal m ⊇ a, then the submodule Ax ⊆ M is such that (Ax)m = 0, and then [3.14] gives Ax = ax,so in particular there is a ∈ a such that x = ax, or (1− a)x = 0.

Deduce thatM = 0 ⇐⇒ Supp(M )∩V (a) =∅

in Spec(A)

.

Recall that the module above is the kernel of the canonical map M → M , so that M = 0 ⇐⇒ M =⋂

m⊇a ker(M →Mm). That in turn means that Mm = 0 for all maximalm⊇ a, or in the language of [3.19], that none of these maximalideals are in Supp(M ). Let p ∈ Spec(A) be contained in a maximal m, so that Sm ⊆ Sp. Then if x ∈ M is annihilatedby an element of Sm, it is a fortiori annihilated by an element of Sp, so by [3.1], if Mm = 0, then also Mp = 0. ThusMm = 0 for all maximal m⊇ a if and only if Mp = 0 for all p ∈V (a), or in other words, Supp(M )∩V (a) =∅.

5 We embarrassingly were unable to make the connection between the above and the fact that p-adic completion is not exact, mostly becausewe were trying to prove surjectivity was not preserved and to find an exact sequence where A, with the alternate topology, occurred as the lastnonzero term. This paragraph adapts Yimu Yin’s solution: http://pitt.edu/~yimuyin/research/AandM/exercises10.pdf

123



4. Let A be a Noetherian ring, a an ideal in A, and A the a-adic completion. For any x ∈ A, let x be the image of x in A.Show that

x not a zero-divisor in A =⇒ x not a zero-divisor in A.

Let x ∈ A not be a zero-divisor, so multiplication by x is injective; the book’s suggested sequence 0→ Ax→ A is

then exact. Taking inverse limits gives an exact sequence 0→ A→ A by (10.3), where the map on A⊆∏

A/an isgiven by ⟨ξn⟩ 7→ ⟨xξn⟩, i.e., by multiplication by x. Thus multiplication by x is injective, so x is not a zero-divisor.

Alternately, but very similarly, one can use the A-algebra structure A→ A to tensor the sequence with A. Since Ais Noetherian, A is flat by (10.14), so the map x⊗id: A⊗AA→A⊗AA is injective. By (2.14.iv) we have an isomorphismφ : A⊗AA→ A, and φ (x ⊗ id) φ−1 : A→ A is multiplication by x.

Does this imply that

A is an integral domain =⇒ A is an integral domain?

Not a priori, for in general not all elements of A are images of elements of A. In fact,6 if we write Aa for the

A/(ab)n+1 ∼ //

A/an+1×A/bn+1

A/(ab)n ∼ // A/an ×A/bn

completion of A at a, then if A is an integral domain and a, b Ã A are coprimeideals, we can show Aab is not an integral domain. Since for each an and bn

remain coprime for n ≥ 1 (see the footnote to [9.8]), by the Chinese RemainderTheorem (1.10), each map A/(ab)n → A/an ×A/bn given by x + (ab)n 7→ ⟨x +an , x + bn⟩ is an isomorphism, and the square diagrams to the right commute,where the vertical maps on the right are ⟨x+an+1, y+bn+1⟩ 7→ ⟨x+an , y+bn⟩.Thus we have an isomorphism between Aab and the inverse limit on the right, which ring consists of coherentsequences

⟨ξn , ηn⟩

n of pairs in∏

(A/an ×A/bn), which correspond to pairs

⟨ξn⟩n , ⟨ηn⟩n

of coherent sequencesin∏

A/an

×∏

A/bn

under the obvious isomorphism, so that Aab ∼= Aa × Ab, and hence is not an integraldomain.

5. Let A be a Noetherian ring and let a, b be ideals in A. If M is any A-module, let Ma, Mb denote its a-adic and b-adiccompletions respectively. If M is finitely generated, prove that (Ma)b ∼=Ma+b.

Since (10.13) tells us Ma = Aa⊗AM , by (2.14) it is enough to prove that Ab⊗AAa ∼= Aa+b, but it is not obvioushow to do this. Instead, we follow the book’s instructions to complete a proof that is unfortunately lengthy whenfully fleshed out.

Following the book’s suggestion, we can a-adically complete the sequences 0→ bm M → M → M/bm M → 0 toget sequences 0→ (bm M )a→ Ma→ (M/bm M )a→ 0, which are exact by (10.12). Using the isomorphisms (10.13)we see the map (bm M )a→Ma is the composition (bm M )a ∼−→Aa⊗Ab

m M Aa⊗AM∼−→Ma. Since the isomorphism

Aa ⊗AM → Aa ⊗AMa → Aa ⊗Aa M → Ma is given by ⟨an⟩ ⊗ x 7→ ⟨an⟩ ⊗ ⟨x⟩ 7→ ⟨an x⟩, it follows that the image of(bm M )a→Ma is bm Ma. Then exactness gives Ma/bm Ma ∼= (M/bm M )a.

Therefore

(Ma)b = lim←− Ma/bm Ma ∼= lim←− (M/bm M )a ∼= lim←−

m

lim←−n

Mbm M

Â

an M + bm Mbm M

,

since (an M +bm M )/bm M is the image of an M under M M/bm M . But by the third isomorphism theorem (2.1.i),the terms on the right-hand side are isomorphic to M/(an M + bm M ) =: Mm, n .

An element of lim←−nMm, n is represented by a sequence ⟨xm, n⟩n of elements xm, n ∈ Mm, n coherent under the

quotient maps Mm, n+1→Mm, n , and an element of lim←−m lim←−n Mm, n is represented by a sequence

⟨xm, n⟩n

m of such

sequences, coherent under the reductions m + 1 7→ m, which act on the nth coordinates xm+1, n (the inverse limitbeing a submodule of

∏

n Mm+1, n) as the quotient maps Mm+1, n → Mm, n . Note that whenever n ≤ n′ and m ≤ m′,the iterated quotient maps Mm′, n′ →Mm′, n →Mm, n and Mm′, n′ →Mm, n′ →Mm, n are equal. Thus an element of thedouble limit is really just an infinite array ⟨xm, n⟩m, n coherent in both directions. Given such a coherent array, eachxp, p uniquely determines all xm, n with m, n ≤ p. Therefore, to specify a coherent array uniquely, one need onlyspecify a coherent diagonal sequence ⟨xp, p⟩ ∈ lim←−p M p, p . This sets up a bijection lim←−m lim←−n Mm, n↔ lim←− M p, p that

obviously preserves the operations and hence is an isomorphism.

6 This is taken, after our own failure, from [KarpukSol].

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Finally, since(a+ b)2n ⊆ an + bn ⊆ (a+ b)n ,

the topology on M induced by the filtration ⟨(an + bn)M ⟩ is the same as that induced by ⟨(a+ b)n M ⟩, so the corre-sponding completions should be isomorphic. But the former is the inverse limit from above, isomorphic to (Ma)b,and the latter is Ma+b.

6. Let A be a Noetherian ring and a an ideal in A. Prove that a is contained in the Jacobson radical of A if and only if everymaximal ideal of A is closed for the a-topology. (A Noetherian topological ring in which the topology is defined by an idealcontained in the Jacobson radical is called a Zariski ring. Examples are local rings and (by (10.15)(iv)) a-adic completions.

This proof oddly requires no use of the Noetherian hypothesis. We show a maximal ideal m is closed in thea-topology if and only if a⊆m. This implies the result because a is contained in all maximal ideals if and only it iscontained in their intersection, the Jacobson radical. Note that a set C ⊆A is closed if and only if each x ∈A\C hasa basic neighborhood x + an disjoint from C . Let mÃA be a maximal ideal.

If a⊆m and x /∈m, then so x + an ⊆ x +m is disjoint from m for all n.If a 6⊆ m, any element of a\m descends to a unit in the field A/m, and so has a multiple x ≡ 1 (mod m), with

x ∈ a. Then xn ∈ an and xn ≡ 1 (mod m), so 1− xn ∈ (1+an)∩m for all n even though 1 /∈m, and m is not closed.

7. Let A be a Noetherian ring, a an ideal of A, and A the a-adic completion. Prove that A is faithfully flat over A (Chapter3, Exercise 16) if and only if A is a Zariski ring (for the a-topology).

We again seem to be able, alarmingly, to prove the result without using the Noetherian hypothesis. By (10.14), Ais flat, and by [3.16.iii], the condition will be met if and only if the extensions mA 6= A for all maximal mÃA. Sincea Noetherian ring A is Zariski if and only if a is in all maximal m, it will be enough to show that for each maximalm we have a⊆m ⇐⇒ mA 6= A ⇐⇒ (IS THIS REALLY THE SAME?) for no element x ∈m is x a unit in A.

If a⊆m, then m/an is a proper ideal of A/an for all n, and so no element of m can become invertible.If a 6⊆ m, then a+m = (1), and there are x ∈ m and a ∈ a with x = 1− a. Inductively, given y2n such that

xy2n = 1− a2n, multiplying both sides by 1+ a2n

gives xy2n (1+ a2n ) = 1− a2n+1, showing x is a unit modulo a2n+1

.Now y2n+1 := y2n (1+ a2n )≡ y2n (mod a2n ), so if we set ym = y2n+1 for 2n < m ≤ 2n+1, we see ⟨yn + an⟩ is an elementof A inverse to x.

Alternately7 (and using the Noetherian hypothesis), if A is faithfully flat, then by [3.16.iii], me 6= A for anymaximal m Ã A, so there is a maximal n Ã A containing it. By [1.5.iv], nc ⊇ mec ⊇ m is maximal, so nc = m.(10.15.iv) says a is in R(A), so a⊆ n. Therefore a⊆ ac ⊆ nc =m, and a is contained in R(A).

The book also has a suggested proof. Since A is flat by (10.14), to prove faithful flatness [3.16.v] says it is enoughto check that the canonical maps M → A⊗AM are injective for all A-modules M . If this fails, some nonzero x ∈ Mis killed, so the composition Ax ,→M → A⊗AM is already not injective. Since A is flat, again by (10.14), A⊗AAx→A⊗AM is injective, so the map Ax→ A⊗AAx is not. Thus we only need to check injectivity for cyclic modules M ;in this case, (10.13) tells us A⊗AM ∼= M , so we are concerned with the maps M → M .

If a is contained in the Jacobson radical of A, then since A is Noetherian and M is finitely generated, by (10.19),the kernel of M → M is 0.

If a is not contained in the Jacobson radical, there is a maximal ideal m /∈V (a); write M = A/m. Since m⊆ A isnot closed by [10.6], it follows that 0 ⊆M is not closed, and so cannot be the kernel of M → M . Alternately, sinceeach an +m = (1), it follows that an M = M for each n, so the kernel is M . Looked at yet another way, by (3.19.v),Supp(M ) =V

Ann(M )

= m, which is disjoint from V (a), so by [10.3], M = 0.

8. Let A be the local ring of the origin in Cn (i.e., the ring of all rational functions f/g ∈C(z1, . . . , zn) with g (0) 6= 0), letB be the ring of power series in z1, . . . , zn which converge in some neighborhood of the origin, and let C be the ring offormal power series in z1, . . . , zn , so that A( B (C . Show that B is a local ring and that its completion for the maximalideal topology is C . Assuming that B is Noetherian, prove that B is A-flat.

To sensibly talk about B , we first must pick a notion of convergence for power series in several variables; we useabsolute convergence, because it is independent of which sequence of partial sums we take the limit of.

7 http://pitt.edu/~yimuyin/research/AandM/exercises10.pdf

125



The surjective ring homomorphism h 7→ h(0) : BC shows the ideal m= (z1, . . . , zn) of power series with zeroconstant term is maximal. To show B is local, we demonstrate a power series f with nonzero constant term is a unit.If the reader is willing to accept that a formal multiplicative inverse of a power series with positive multiradius ofconvergence likewise has positive multiradius of convergence, then we need only produce an f −1 ∈ C ; and we cando this by an induction starting at n = 0, for by [1.5.i], f ∈C[[z1, . . . , zn , w]] is a unit just if f (z, 0) ∈C[[z1, . . . , zn]]is a unit. Otherwise, please consult the footnote.8 This also shows that A⊆ B , since any g ∈C[z] with g (0) 6= 0 hasan inverse in B . But A( B , since for example exp(z) /∈A. We see B (C since

∑

j !z j1 ∈C but j !|ζ j | →∞ for ζ 6= 0.

To compute B , write n=m∩A and note that the inclusion A ,→ B induces isomorphisms A/nm ∼−→ B/mm for allm ∈N, basically because the truncations of power series below a given total degree are just polynomials. Since theseisomorphisms respect the quotient maps A/nm+1 A/nm , we have an isomorphism of inverse systems, so B ∼= A,which, as pointed out in the proof of (10.27), is C .

Since A and B are local rings with topologies defined by their maximal ideals (which are their Jacobson radicals),and we are told B is Noetherian ([7.4.ii] contains a proof in the n = 1 case; can we do an induction?), they are Zariskirings. By [10.7], C = B = A is faithfully flat over A and B , so it follows from [3.17] that B is A-flat.

9. Let A be a local ring, m its maximal ideal. Assume that A is m-adically complete. For any polynomial f (x) ∈ A[x],let f (x) ∈ (A/m)[x] denote its reduction mod. m. Prove Hensel’s lemma: if f (x) is monic of degree n and if there existcoprime monic polynomials g (x), h(x) ∈ (A/m)[x] of degrees r, n− r with f (x) = g (x)h(x), then we can lift g (x), h(x)back to monic polynomials g (x), h(x) ∈A[x] such that f (x) = g (x)h(x).

There is a more general version stated as [Eisenbud, Thm. 7.18], proved there in two exercises whose extensivehints we follow.

First, we show that given a ring B and coprime, monic p, q ∈ B[x] with deg p = r, any c ∈ B[x] admits anexpression c = ap + b q with deg b < r. (In the case q = 1, this is the division algorithm. This implies the book’ssuggested lemma, for if deg q = n− r and deg c ≤ n, we have deg b q < n, forcing deg a ≤ n− r.) Indeed, B[x]/(p)is generated by q since (p, q) = B[x], so there is b ∈ B[x], such that b q = c in B[x]/(p). Since p is monic ofdegree r , the elements 1, x, . . . , x r−1 freely generate B[x]/(p) as a B -module, so we may assume deg b < r . Since qis monic, this choice of b is unique. Now c ≡ b q

mod (p)

, and p, being monic, is by [1.2.ii] not a zero-divisor, soap = c − b q has a unique solution a ∈ B[x]. 9, 10

8 This is adapted from a proof where n = 1 in [Lang, Ch. II, Thm. 3.3]. As convergence is unaffected by taking constant multiples, wemay assume without loss of generality that f (0) = 1. Here we set up some notation. We will work in the rings C[[z, w]] = C[[z1, . . . , zn , w]]and C[[w]]. If α = ⟨α1, . . . , αn⟩ ∈ Nn is a multi-index, xα =

∏

xαmm , whether xm = zm or xm ∈ C. If |a j | ≤ r j for a j ∈ C and r j ≥ 0, write

∑

a j w j ≺∑

r j w j in C[[w]]; this is a transitive relation. The following fact is helpful: if f , g ∈C[[w]] andω ∈C are such that f ≺ g and g (ω)converges, then f (ω) converges.

To see f −1 converges in a neighborhood of the origin, set g = 1− f ; then formally, f −1 = (1− g )−1 =∑

k g k , using the geometric series.Write g =

∑

a j w j ∈ C[[z, w]] for a j ∈ C[[z]]. Since g converges absolutely some neighborhood of the origin, we can find a closed polydisk

⟨ζ ,ω⟩ : ζm ≤ εm , ω ≤ δ on which g converges absolutely. In particular, the a j converge absolutely on D = ζ : ζm ≤ εm. If we write

a j =∑

cαzα for cα ∈ C, then their maximal values on D are the finite numbers r j =∑

|cα|εα, and so to show f −1(ζ ,ω) converges for all

ζ ∈ D it suffices to show f −1(r,ω) := f −1(r1, . . . , rn ,ω) converges. If we had r j ≥ δ− j for infinitely many j , the series g (r, δ) would notconverge; as we know it does, r j <δ

− j for all but finitely many j , and thus there is a constant R≥ δ−1 such that r j < R j for all j . We now have

g (r, w)≺∑∞

j=1 R j w j = Rw1−Rw , so

f −1(r, w)≺∞∑

k=0

g (r, w)k ≺∞∑

k=0

Rw1−Rw

k=

1

1− Rw1−Rw

=1−Rw1− 2Rw

= (1−Rw)∞∑

k=0

(2Rw)k ≺ (1+Rw)∞∑

k=0

(2Rw)k ,

showing f −1 converges on D ×ω : |ω|< 1/2R.9 There is actually a mistake here ([EisenEmail]), as the author does not require q to be monic. We need q to not be a zero-divisor for the

uniqueness claim. This will also be the case if B[x]/(p) is an integral domain, but we cannot make any guarantees on p in the rest of the proof.10 Here is a proof of the book’s lemma that if monic p, q ∈ B[x] of respective degrees r, n − r are coprime, then for any polynomial c of

degree ≤ n there are a, b ∈ B[x], of respective degrees ≤ n− r, r , such that ap + b q = c . Since (p, q) = (1), this is obviously possible if we dropthe restriction on degrees. If a = a j x j + (deg < j ) and b = bm x m + (deg < m) with a j , bm 6= 0, and j > n − r , we will polynomials a′, b ′ ofdegrees respectively < j , m such that again a′p + b ′q = c ; applying this repeatedly eventually achieves the desired restriction on degrees. Sincedeg c ≤ n < j + r , it follows that the leading term a j x j+r of ap cancels the leading term bm x m+n−r , so a j = −bm and j + r = m + n − r , or

j − (n− r ) = m− r . (This also shows r < m.) If we let a′ = a− a j x j−(n−r )q and b ′ = b − bm x m−r p, then deg a′ < j and deg b ′ < m, and

a′p + b ′q = (ap + b q)− (a j x j−(n−r )+ bm x m−r )pq = c .

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Second, we claim that if b⊆R(A) is an ideal of A and we are given g , h ∈A[x]with g monic, then if ( g , h) = (1)in (A/b)[x], we have (g , h) = A[x]. Indeed, if M = A[x]/(g ), and N = hM , then since bA[x] + (g , h) = A[x],we have bM +N = M . Since M is finitely generated by 1, x, . . . , xdeg g−1, Nakayama’s Lemma (2.7) applies to showhM =M , so that (g , h) = (1) in A[x].

Now we prove the result. We need only assume that A is complete with respect to some ideal a; we do notnecessarily need amaximal or A local. We inductively construct sequences ⟨gk⟩ and ⟨hk⟩ of monic polynomials of theright degrees in A[x]with f −gk hk ≡ 0 (mod ak ) and gk ≡ g j , hk ≡ h j (mod a j ) for j < k. Coefficient by coefficient,the gk and the hk will form Cauchy sequences with respect to the a-topology, and so converge to unique limitsg , h ∈A[x] with g ≡ gk , h ≡ hk (mod ak ) for all k ≥ 1. We will then have f − g h = ( f − gk hk )+ (gk hk − g h)≡ 0(mod ak ) for all k. Since A is a-adically complete,

⋂

ak = 0, so f = g h, as hoped.If g1, h1 are lifts of the appropriate degrees of g , h to A[x], by assumption, f − g1h1 ≡ 0 (mod a) (really meaning

modulo a[x]). The Eisenbud version of the induction step follows; the one from the book is in the footnote.11

Suppose inductively we have found gk , hk ∈ A[x] such that gk ≡ g j , hk ≡ h j (mod a2 j−1) for all j < k and f −gk hk = c ≡ 0 (mod a2k−1). Since by (10.15.iv), a2k−1 ⊆ R(A), it follows from the second claim, with b = a2k−1

, that(gk , hk ) = A[x]. By the first claim, with B = A, p = gk , q = hk , there are unique a, b ∈ A[x] with deg b < r andagk+bhk = c . Descending to A/a2k−1[x], where we take B =A/a2k−1

in the first claim and use uniqueness, we see thatsince c = 0 we also have a = b = 0, or a, b ∈ a2k−1[x]. Set gk+1 = gk+b ∈ gk+a

2k [x] and hk+1 = hk+a ∈ hk+a2k [x],

sof − gk+1hk+1 = ( f − gk hk )− (agk + bhk )− ab = c − c + ab = ab ∈ a2k

[x].

10. i) With the notation of Exercise 9, deduce from Hensel’s lemma that if f (x) has a simple root α ∈ A/m, then f (x) has asimple root a ∈A such that α= a mod m.

If α is a simple root of f , we have a factorization f = (x −α) g , where g is coprime to x −α. Since we assume fis monic, so is g . By [10.9], there exist monic lifts h of x−α and g of g to A[x]; since deg h = 1, we must be able towrite h = x − a for some a ∈A, and a 7→ α under A→A/m. Since x − a is irreducible, if it divided g , then (x −α)2

would divide f , contrary to assumption, so a is a simple root of f .

ii) Show that 2 is a square in the ring of 7-adic integers.2 is be a square if and only if x2− 2 ∈ Z7[x] has a solution, if and only if it splits into linear factors. x2− 2 has

simple roots ±3 in Z7/7Z7∼=Z/7Z, so by i), 2 has two square roots in Z7.

iii) Let f (x, y) ∈ k[x, y], where k is a field, and assume that f (0, y) has y = a0 as a simple root. Prove that there exists aformal power series y(x) =

∑∞n=0 an xn such that f

x, y(x)

= 0.(This gives the “analytic branch” of the curve f = 0 through the point (0, a0).)

Write A = k[[x]] and m = (x), so A/m = k. Considering f (x, y) as an element of A[y] ) k[x][y] = k[x, y],the image of f in k[y] is f (0, y), which by assumption has a simple root a0 ∈ k. By i), f (x, y) has a simple rooty(x) ∈ k[[x]] with constant term a0.

11. Show that the converse of (10.26) is false, even if we assume that A is local and that A is a finitely-generated A-module.Since (10.26) says that completing a Noetherian ring A with respect to an ideal a produces a Noetherian A, the

converse would presumably be that if the a-completion A of a ring A is Noetherian, then A was already Noetherian.It then falls to us to find a non-Noetherian local ring with finitely generated, Noetherian completion.

11 Let gk , hk ∈A[x] be given. Since gk = g , hk = h, we have ( gk , hk ) = (1) in (A/a)[x], so there are m j ∈ ak such that f − gk hk =∑n

j=0 m j x j ,

and for 0≤ j ≤ n there exist a j , b j ∈A[x] of degrees≤ n− r, r such that a j gk+ b j hk = x j in (A/a)[x]. Then there are r j ∈ a[x]with deg r j ≤ nsuch that a j gk + b j hk + r j = x j in A[x] (this paraphrases the solution in [PapaSol]; I had not seen the necessity of working coefficient bycoefficient and gotten stuck). We then can write

f − gk hk =∑

m j (a j gk + b j hk + r j ) = gk

∑

m j a j + hk

∑

m j b j +∑

m j r j .

Now gk+1 = gk +∑

m j b j ∈ gk + ak+1[x] and hk+1 = hk +∑

m j a j ∈ hk + ak+1[x] satisfy the degree restrictions, and

f − gk+1 hk+1 = ( f − gk hk )− gk

∑

m j a j − hk

∑

m j b j −∑

j

∑

`

m j m`a j b` =∑

m j r j −∑

j

∑

`

m j m`a j b` ∈ ak+1[x].

127


Following the book’s hint, let A = C∞0 (R) be the set of germs [ f ] at 0 ∈ R of C∞ real-valued functions f.The homomorphism [ f ] 7→ f (0) surjects onto the field R, showing the functions vanishing at 0 form a maximalideal a. To see this is the only maximal ideal, suppose f represents an [ f ] /∈ a; then by continuity f 6= 0 on someneighborhood of 0, so f locally admits a multiplicative inverse g , and [ f ][g ] = [1], showing [ f ] is a unit.

Write bn :=

[ f ] ∈ A : f ( j )(0) = 0 for 0 ≤ j < n

; we want to show an = bn = (xn). Suppose inductively that

an ⊆ bn . Then if [ f ] ∈ an+1, we can write f as a sum of elements g h for [g ] ∈ a and [h] ∈ an . By the generalizedproduct rule, (g h)(n) =

∑nj=0

nj

g ( j )h (n− j ); since g (0) = 0, and h (n− j )(0) = 0 for j ≥ 1, we see f (n)(0) = 0, so[ f ] ∈ bn+1. For the reverse inclusion, we use Taylor’s theorem with remainder: for any open interval U ⊆ R andf ∈ C∞(U ), we can write f (x) =

∑n−1j=0

1j ! f ( j )(0)x j + gn(x)x

n on U , where gn ∈ C∞(U ) and gn(0) =1n! f (n)(0).12

Thus bn ⊆ (xn)⊆ an .Since xn gn ∈ an , and any polynomial is its own Maclaurin series, truncations of Maclaurin series yield isomor-

phisms A/an→R[x]/(xn) compatible with the quotient homomorphisms given by n+1 7→ n. By Example 1 on p.105, A∼=R[[x]], which is Noetherian by (7.5*) becauseR is a field. However, by (10.18), since 0 6= e−1/x2 ∈

⋂

an (itsMaclaurin series is 0), A is not Noetherian. By Borel’s theorem that every power series is the Taylor series of a C∞

function, A→ A is surjective, so that A is finitely generated.

12. If A is Noetherian, then A[[x1, . . . , xn]] is a faithfully flat A-algebra.By [2.5], A→ A[x1, . . . , xn] is flat, and by (10.14) and our proof above that the latter is the completion of the

former, A[x1, . . . , xn] → A[[x1, . . . , xn]] is flat, so [2.8.ii] says A→ A[[x1, . . . , xn]] is flat. For any a Ã A we haveae = a+ a · (x1, . . . , xn), so that aec = a; thus, by [3.16.i], A[[x1, . . . , xn]] is faithfully flat over A.

12 For a proof, note that by the chain rule, ddt f (tx) = x f ′(tx). Integrating both sides from 0 to 1 and using the fundamental theorem of calculus

gives f (x)− f (0) = x∫ 1

0 f ′(tx)dt . Write g (x) =∫ 1

0 f ′(tx)dt ; then g (0) =∫ 1

0 f ′(0)dt = f ′(0). Thus f (x) = f (0) + x g (x) for a C∞ function gwith g (0) = f ′(0). See [Tu, Lemma 1.4] for a generalization of this result to real-valued functions on open subsets ofRn , star-shaped with respectto some point (specialized to 0 for us).

Applying the above to g and iterating, we get expressions f (x) = f (0)+∑n−1

j=1 g j (0)xj + gn(x)x

n , so f (n)(0) = n!gn(0) for all n.

This differs slightly from the usual form of the theorem, which assumes only that f is n times differentiable at 0 and gets a remainder termhn−1(x)x

n−1 with limx→0 hn−1(x) = 0 instead of our gn(x)xn ; see e.g. [WPTaylor].

128

Chapter 11: Dimension Theory

(1− t )−d =∞∑

k=0

d + k − 1d − 1

t k . (p. 117)

(1−t )−1 = 1+t+t 2+· · · , so we want to check that the coefficient of t k in its d th power isd+k−1

d−1

. This coefficientis the number of possible ways of forming a product

∏dj=1 t k j with

∑dj=1 k j = k, which is the number of ordered

partitions of a row of k objects into d groups. This is the same as the number of ways of inserting d−1 dividers intothe row of k objects, or of choosing d − 1 objects out of k + d − 1 to serve as dividers, namely

d+k−1d−1

.

Example. Let A=A0[x1, . . . , xs ], where A0 is an Artin ring and the xi are independent indeterminates. Then An is a freeA0-module generated by the monomials x m1

1 · · · xmss where

∑

mi = n; there ares+n−1

s−1

of these, hence P (A, t ) = (1− t )−s .(p. 118)

The coefficient of t n in P (A, t ) is l (An). Since An∼= A(

s+n−1s−1 )

0 as an A0-module, it follows l (An) =s+n−1

s−1

l (A0).Then by the expression above, P (A, t ) = l (A0)(1− t )−s . This is not the expression that the book gives unless l (A0) =1. The degree d (A) = d

Gm(A)

as defined on p. 119 is unaffected by this change, and hence so is the dimension n ofk[x1, . . . , xn](x1, ..., xn )

in the example on p. 121 and d

Gq(A)

in the proof of (11.20).

Given a polynomial f (x) ∈Z[x], the sum g (n) =∑n

j=0 f ( j ) is a polynomial in n.* (p. 119)

We can write f (n) =∑

k ak nk , so it will be enough to show that for each n the function gk (n) =∑n

j=0 j k is apolynomial.1 Note that g0(n) = n + 1. Suppose inductively that g j (n) is a polynomial for j ≤ k. By the binomial

theorem we have (m + 1)k+1 − mk+1 =∑k

j=0

k+1j

m j . Summing both sides from m = 0 to n gives (n + 1)k+1 =∑k

j=0

k+1j

g j (n), and rearranging, we see gk (n) = (n+ 1)k+1−∑k−1

j=0

k+1j

g j (n) is a polynomial in n.

If 0→N ,→M →M ′→ 0 is exact and qÃA, then 0→N/(N ∩ qn M )→M/qn M →M ′/qn M ′→ 0 is exact.* (p. 120)This is a special case of a similar result in the beginning notes to Ch. 10.

EXERCISES1. Let f ∈ k[x1, . . . , xn] be an irreducible polynomial over an algebraically closed field k. A point P on the variety f (x) = 0

is non-singular ⇐⇒ not all the partial derivatives.f/.xi vanish at P. Let A = k[x1, . . . , xn]/( f ), and let m be themaximal ideal of A corresponding to the point P. Prove that P is non-singular ⇐⇒ Am is a regular local ring.

Write k[x] = k[x1, . . . , xn] and let P = ⟨a1, . . . , an⟩, so that m is the image in A of mP = (x1− a1, . . . , xn − an)Ãk[x]. Since f (P ) = 0, we have f ∈ mP , so there are p j ∈ k[x] (possibly zero) such that f =

∑

(x j − a j )p j . Then

.f/.xi = pi + (xi − ai )(.pi/.xi ) +∑

j 6=i (x j − a j ).p j/.xi . All terms but possibly pi are in mP ; so it follows that P is asingular point of the variety f (x) = 0 ⇐⇒ all the.f/.xi vanish at P ⇐⇒ each pi ∈mP ⇐⇒ f ∈m2

P .Now we work to rephrase regularity of Am in terms of f . Since A= k[x]/( f ) and the quotient map k[x]→ A

takes SmP→ Sm, by (3.4.iii) and [3.4], we have k[x]mP

/( f )mP∼=

k[x]/( f )

m= Am. Since dim k[x]mP

= n, by(11.18), dim Am = n− 1. By the third isomorphism theorem (2.1.i) and (3.4.iii), k ∼= k[x]/mP

∼= A/m∼= Am/mAm,so, using (3.4.iii) again, Am is a regular local ring just if dimk (m/m

2) = dimk (mAm/m2Am) = n−1. Nowm=mP/( f ),

and m2 is the image of m2P under k[x]→A, which is

m2P +( f )

/( f ), so that by (2.1.i) again, m/m2 = mP( f )

.

m2P+( f )( f )

∼=

1 adapted from http://mathforum.org/library/drmath/view/56920.html

129

http://mathforum.org/library/drmath/view/56920.html

Ex. 11.2 Chapter 11: Dimension Theory

mP/

m2P + ( f )

. If f ∈ m2P , then this is mP/m

2P , which has dimension n, so Am is not regular. Otherwise m2

P + ( f )strictly contains m2

P , so dimk (m/m2)< n. But by (11.15), dimk (m/m

2)≥ n− 1.

2. In (11.21) assume that A is complete. Prove that the homomorphism k[[t1, . . . , td ]]→ A given by ti 7→ xi (1≤ i ≤ d ) isinjective and that A is a finitely-generated module over k[[t1, . . . , td ]].

The first thing to check is that this homomorphism is well defined. If n = (t1, . . . , td ) Ã k[t ] := k[t1, . . . , td ],then the nn form a neighborhood basis of 0 in k[t ], and the map k[t ]→ A given by ti 7→ xi sends nn → qn ⊆ mn .Thus a Cauchy sequence in k[t ] is sent to a Cauchy sequence in A, which converges to a point of A by completeness,so since k[[t ]] := k[[t1, . . . , td ]] is the completion of k[t ] with respect to n, we have a well defined map k[[t ]]→Aas asserted, making A a k[[t ]]-module.

For injectivity, let p(t ) ∈ k[[t ]] be in the kernel. Writing q = (x1, . . . , xd ), by (7.16.iii), p(x) = 0 ⇐⇒ p(x) ∈qn for every n. Write pn(t ) for the nth homogeneous component of p(t ); since p0(x) = 0, it follows p0(t ) = 0.Inductively suppose p j (t ) = 0 for all j ≤ n. Now p(x)− pn(x) ∈ qn+1, and p(x) = 0 ∈ qn+1, so pn(x) ∈ qn+1 By(11.20), the coefficients of pn(t ) are in k ∩m= 0. Thus each pn(t ) = 0, so p(t ) = 0.

Now ⟨qn⟩ is a n-filtration of A, k[[t ]] is complete, and A is Hausdorff (since complete) with respect to the q-topology (which by (7.16.iii) is the m-topology), so by (10.24), A will be a finitely-generated k[[t ]]-module if Gq(A)is a finitely-generated Gn

k[[t ]]

-module. By (10.22.ii), Gn

k[[t ]] ∼= Gn

k[t ]

and the latter is isomorphic to k[t ]under ti 7→ ti . But the map k[t ]→Gq(A) taking ti 7→ xi is a quotient map.

3. Extend (11.25) to non-algebraically-closed fields.By Noether normalization [5.16], if d = dim V , there are algebraically independent x1, . . . , xd ∈A(V ) such that

A(V ) is integral over the polynomial subring B = k[x1, . . . , xd ]. Since B is a UFD, as noted on p. 63, it is integrallyclosed. Following the book’s hint, note that k is integral over k ( B , and trivially the xi are integral over B , so thatby (5.3) the ring C = k[x1, . . . , xd ] is integral over B . As V is an irreducible variety, A(V ) is an integral domain,so that (11.26) applies to the inclusion B ⊆ A(V ): for any maximal ideal m Ã A(V ), then, dim A(V )m = dim Bmc ,where mc is maximal by (5.8). It remains to show that dim Bn = d for all maximal ideals n Ã B . By (5.10), since Cis integral over B , there is a prime of C lying over n, and by (1.4), there is a maximal ideal q Ã C containing thatprime, whose contraction to B is then a prime ideal containing n, which must be n itself. As (11.26) also applies tothe inclusion B ⊆C , we have dim Bn = dim Cq. But we already established in (11.25) that dim Cq = d .

4. An example of a Noetherian domain of infinite dimension (Nagata). Let k be a field and let A= k[x1, x2, . . . , xn , . . .]be a polynomial ring over k in a countably infinite set of indeterminates. Let m1, m2, . . . be an increasing [or even justunbounded] sequence of positive integers such that mi+1 − mi > mi − mi−1 for all i > 1. [Actually, set m1 = 0.] Letpi = (xmi+1, . . . , xmi+1

) and let S be the complement in A of the union of the ideals pi .Each pi is a prime ideal and therefore the set S is multiplicatively closed. The ring S−1A is Noetherian by Chapter 7,

Exercise 9. Each S−1pi has height equal to mi+1−mi , hence dim S−1A=∞.Note that the complement S of the union of a set P of primes in A is multiplicatively closed, as follows: x, y ∈

S ⇐⇒ ∀p ∈ P (x, y /∈ p) ⇐⇒ ∀p ∈ P (xy /∈ p) ⇐⇒ xy ∈ S.The prime ideals of A that persist in S−1A are by (3.11.iv) those that don’t meet S, so the maximal ideals of S−1A

are S−1p for prime p Ã A maximal with respect to not meeting S. Suppose a Ã A doesn’t meet S, so it is containedin the union of the pi ; we claim it is contained in some pi . Given (a1, . . . , a`)⊆ a, the a j only involve finitely manyindeterminates, so for some n we have (a1, . . . , a`) ⊆

⋃ni=1 pi . By (1.11.i), (a1, . . . , a`) ⊆ pi for some i ≤ n. If we

append a`+1 ∈ a, we again have an n′ ∈N such that (a1, . . . , a`, a`+1)⊆ pi for some i ≤ n′; but n′ ≤ n, since we don’thave a1 ∈ pi for i > n by assumption. Thus appending a generator can only decrease our collection of candidate pi .If for any b ∈ a we have b /∈ pi , we can choose a`+1 = b and pick a new pi with i ≤ n. Since there are only finitelymany of these, this is eventually no longer possible, and then we are done.

Thus the maximal ideals of S−1A are the S−1pi . We claim the localizations with respect to these are Noetherian.Without loss of generality, take i = 1. Note

S−1A\S−1p1

−1 =

S−1(A\p1)−1 =

S−1Sp1

−1 = S−1p1

S = S−1p1

sinceS =A\

⋃

p j ⊆A\p1 = Sp1and Sp1

is multiplicatively closed. Thus the localization

(S−1A)S−1p1=

S−1A\S−1p1

−1(S−1A) = S−1p1

S−1A= S−1p1

A=Ap1.

If we let K be the field k(xm2+1, . . . , xm2+n , . . .), then Ap1=K[x1, . . . , xm2

](x1, ..., xm2) as a subset of the field of fractions

of A. Since K[x1, . . . , xm2] is Noetherian by the Hilbert Basis Theorem (7.6), Ap1

is Noetherian by (7.4). By (3.13),the height of S−1p1 is dim (S−1A)S−1p1

= dim Ap1= dim K[x1, . . . , xm2

](x1, ..., xm2); this is m2 by the example on p. 121.

130

Chapter 11: Dimension Theory Ex. 11.5

Since each nonzero a ∈A only uses finitely many indeterminates, it can only be in finitely many pi , and so eachnonzero a/s ∈ S−1A can only be in finitely many maximal ideals S−1pi . This and the fact that the (S−1A)S−1pi

areNoetherian are the hypotheses of [7.9], which tells us S−1A is Noetherian.

5. Reformulate (11.1) in terms of the Grothendieck group K(A0) (Chapter 7, Exercise 25).We recall the hypotheses and definitions. A =

⊕

An is a Noetherian graded ring, generated as an algebra overits summand A0 by finitely many homogeneous elements x j , j = 1, . . . , r , of respective degrees k j > 0. M =

⊕

Mnis a finitely generated graded A-module, so that each Mn is a finitely generated A0-module. λ is an additive functionfrom the the class of finitely generated A0-modules to Z. The Poincaré series of M with respect to λ is Pλ(M , t ) =∑∞

n=0 λ(Mn)tn ∈Z[[t ]]. (11.1) states that, with q =

∏rj=1(1− t k j ) and the standard definition for reciprocals of

power series, Pλ(M , t ) ∈ q−1Z[t ](Z[[t ]].I feel my attempts to say something meaningful about this situation in terms of K(A0) have come up a little short,

but here goes.Write Fgr(A) for the category of finitely generated graded A-modules and degree-preserving A-module homo-

morphisms. Define the graded Grothendieck group Kgr(A) of A from theFgr(A) using the same process by which wedefined the original K -group: form the free abelian group on the set of isomorphism classes ofFgr(A) and take thequotient by the subgroup generated by [N ]− [M ]+ [P ] for all short exact sequences 0→N →M → P → 0 (wherenow the isomorphisms defining the classes and the maps in the sequences are degree-preserving). Write γgr(M ) forthe class of M in Kgr(A) and γ (Mn) for the class of Mn in K(A0). As a degree-preserving homomorphism of graded A-modules induces an A0-module homomorphism on each component, there is a natural map Φ : Kgr(A)→K(A0)[[t ]]given by γgr(M ) 7→

∑

γ (Mn)tn , where K(A0)[[t ]] is just an additive group. There does not seem to be a reason to

expect Φ to be either surjective or injective.Since λ is additive, it induces a homomorphism λ0 : K(A0)→ Z as in [7.26.i], which we can apply to each com-

ponent in K(A0)[[t ]]; call this process Λ0]. Then we can factor Pλ(−, t ) as

Fgr(A)γgr→Kgr(A)

Φ→K(A0)[[t ]]Λ0−→Z[[t ]] :

M 7→ γgr(M ) 7→∑

γ (Mn)tn 7→

∑

(λ0 γ )(Mn)tn =

∑

λ(Mn)tn .

From this one can see that the original additive functionλ doesn’t matter so much as the associatedλ0 ∈Hom

K(A0), Z).Thus im(Φ γgr)( K(A0)[[t ]] is a collection of “universal Poincaré series” for finitely generated graded A-modules,each of which, when subjected to any λ0 ∈Hom

K(A0), Z), produces an element of q−1Z[t ](Z[[t ]].2 Thus, finally,the best we can do in the general case is to say that the Poincaré series yields a bilinear map

Q : Hom

K(A0), Z

×Kgr(A)→Z[[t ]]

with imQ = q−1Z[t ]. This looks a bit different than the original, but is not much more interesting.3

6. Let A be a ring (not necessarily Noetherian). Prove that

1+ dim A≤ dim A[x]≤ 1+ 2 dim A.

If p0 ( p1 ( · · ·( pr is any ascending chain of length r in Spec(A), then p0[x]( p1[x]( · · ·( pr [x]( pr + (x)is an ascending chain of length r + 1 in Spec

A[x]

; the last ideal is prime since it is the kernel of A[x]A/pr andthe others by [2.7]. Thus 1+ dim A≤ dim A[x].

By [3.21.iv], for p ∈ Spec(A) we have a homeomorphism between the set of primes of A[x] lying over p andSpec

k(p)⊗AA[x]

, where k(p) is the field Ap/pAp. By [2.6], k(p)⊗AA[x]∼= k(p)[x]; but dim k(p)[x] = 1, becauseany nonzero prime is maximal, so a chain of primes of A[x] over a given p has length no more than one, and hencecontains at most two primes. Thus a chain of length r in Spec(A), containing r + 1 primes, is the contraction ofa chain of at most 2r + 2 primes of A[x], which has length 2r + 1. Taking suprema over chains in Spec(A) givesdim A[x]≤ 1+ 2 dim A.

2 One would like to lift the result about the image up to K(A0)[[t ]], and say something like “im(Φ γgr) ⊆ q−1K(A0)[t ],” but since K(A0)doesn’t usually have a ring structure, multiplication and hence q−1 have no obvious meaning in K(A0)[[t ]].

3 With more restrictive hypotheses, we can say more; see http://math.stackexchange.com/questions/217612/exercise-11-5-from-atiyah-macdonald-hilbert-serre-theorem-and-grothendieck-grou and the articles by William Smokelinked therein.

131

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http://math.stackexchange.com/questions/217612/exercise-11-5-from-atiyah-macdonald-hilbert-serre-theorem-and-grothendieck-grou

Ex. 11.7 Chapter 11: Dimension Theory

7. Let A be a Noetherian ring. Then

dim A[x] = 1+ dim A,

and hence, by induction on n,dim A[x1, . . . , xn] = n+ dim A.

By [11.6], we have dim A[x]≥ 1+ dim A.For the other direction, it will be enough to show that height P≤ height p+ 1 whenever P is a prime of A[x]

and p = Pc Ã A. If we form rings of fractions of both rings with respect to A\p to get Pp Ã Ap[x] lying overpp Ã Ap, we see by (3.11.iv) that these have the same heights as P and p, respectively, so we may assume A is localwith maximal ideal p.

Following the book’s hint, we first show height p[x] = height p. If height p= m, then since A is a local Noethe-rian ring, there is by (11.13) a p-primary ideal q of A generated by m elements. q[x] is p[x]-primary by [4.7.iii], andis generated over A[x] by the m generators of q, so by (11.16), height p[x]≤ m. We proved m ≤ height p[x], on theother hand, in [11.6].

Now we show, by induction on height p, that height P ≤ height p+ 1. For height p = 0, the result is againimplied by the proof of [11.6]. Suppose that height p = m and the result holds for primes of height < m. To showheight P ≤ m + 1, we need to see height Q ≤ m for each prime Q ( P. If Qc ( p, then height Qc < m, soheight Q ≤ m by induction. If Qc = p, then p[x] ⊆ Q ( P, and so, recalling from our proof of [11.6] that thelongest chain of primes in A[x] lying over p contains two primes, we see Q= p[x], whose height is m.

132

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